On the way to school, Karen became fixated on the puzzle game on her phone!

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:
• row x, (1 ≤ x ≤ n) describing a move of the form “choose the x-th row”.
• col x, (1 ≤ x ≤ m) describing a move of the form “choose the x-th column”.

If there are multiple optimal solutions, output any one of them.

Examples

Input
3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1

Output
4
row 1
row 1
col 4
row 3

Input
3 3
0 0 0
0 1 0
0 0 0

Output
-1

這道題的大意就是一個m*n的二維陣列初始為0，問最少經過幾次操作（每次操作可以每一行或每一列對應的每個數加1）得到給出的二維陣列並輸出操作過程（多種可能性輸出任意一種即可）。

如果列數大於行數，說明是一個扁的二維陣列，找到每一行的最小值並記錄（用於對行的輸出），把該行所有數減去這個最小值；現在每一行的每個數之和應該相等，[若不相等輸出-1（證明不能操作成功）]只看第一行進行對列的輸出即可。
如果行數大於列數，說明是一個瘦高的二維陣列，進行類似的操作即可。

#include<iostream>
#include<string.h>
using namespace std;
int grid[510][510];
int mmin[510],sum[510];
int main(){
    int col,roo;
    while(cin>>roo>>col){
        memset(grid,0,sizeof(grid));
        memset(sum,0,sizeof(sum));
        for(int i=0;i<510;i++)
            mmin[i]=9999;
        for(int i=0;i<roo;i++)
            for(int j=0;j<col;j++)
                cin>>grid[i][j];

        int chose=max(roo,col);
        int ans=0;
        if(chose==col){
            for(int i=0;i<roo;i++)
                for(int j=0;j<col;j++)
                    if(grid[i][j]<mmin[i]) mmin[i]=grid[i][j];
            for(int i=0;i<roo;i++){
                ans=ans+mmin[i];
                for(int j=0;j<col;j++){
                        grid[i][j]=grid[i][j]-mmin[i];
                        sum[i]=sum[i]+grid[i][j];
                    }
            }
            for(int i=1;i<roo;i++)          
                if(sum[i]!=sum[i-1]){ 
                    ans=-1;break;           
            }
            if(ans==-1){
                cout<<"-1"<<endl;
                continue;
            }            
            for(int j=0;j<col;j++){
                if(grid[0][j]!=0){
                    ans=ans+grid[0][j];
                }
            }
            cout<<ans<<endl;
            for(int i=0;i<roo;i++){
                for(int j=0;j<mmin[i];j++)
                    cout<<"row "<<i+1<<endl;
            }
            for(int j=0;j<col;j++){
                if(grid[0][j]!=0){              
                    for(int i=0;i<grid[0][j];i++){
                        cout<<"col "<<j+1<<endl;
                    }
                }
            }
        }
        else{

            for(int i=0;i<col;i++)
                for(int j=0;j<roo;j++)
                    if(grid[j][i]<mmin[i]) mmin[i]=grid[j][i];
            for(int i=0;i<col;i++){
                ans=ans+mmin[i];
                for(int j=0;j<roo;j++){
                        grid[j][i]=grid[j][i]-mmin[i];
                        sum[i]=sum[i]+grid[j][i];
                    }
            }
            for(int i=1;i<col;i++)          
                if(sum[i]!=sum[i-1]){ 
                    ans=-1;break;           
            }
            if(ans==-1){
                cout<<"-1"<<endl;
                continue;
            }            
            for(int j=0;j<roo;j++){
                if(grid[j][0]!=0){
                    ans=ans+grid[j][0];
                }
            }
            cout<<ans<<endl;
            for(int i=0;i<col;i++){
                for(int j=0;j<mmin[i];j++)
                    cout<<"col "<<i+1<<endl;
            }
            for(int j=0;j<roo;j++){
                if(grid[j][0]!=0){              
                    for(int i=0;i<grid[j][0];i++){
                        cout<<"row "<<j+1<<endl;
                    }
                }
            }
        }           
    }

    return 0;
}

然而這種寫法程式碼冗長，而且寫迴圈控制位置的時候容易出錯，於是想到了二維陣列的轉置。

#include<iostream>
#include<string.h>
using namespace std;
int grid[510][510];
int temp[510][510]; 
int mmin[510],sum[510];
int main(){
    int col,roo;
    while(cin>>roo>>col){
        memset(grid,0,sizeof(grid));
        memset(sum,0,sizeof(sum));
        for(int i=0;i<510;i++)
            mmin[i]=9999;
        for(int i=0;i<roo;i++)
            for(int j=0;j<col;j++)
                cin>>temp[i][j];

        int chose=max(roo,col);
        int ans=0;
        int det=0;
        if(chose==roo){
            //轉置並交換行列數 併為轉置變數賦值為1 
            for(int i=0;i<roo;i++)
                for(int j=0;j<col;j++)
                    grid[j][i]=temp[i][j];   
            int tttemp=col;col=roo;roo=tttemp;  
            det=1;

        }
        else 
            for(int i=0;i<roo;i++)
                for(int j=0;j<col;j++)
                    grid[i][j]=temp[i][j];//直接賦值過去 

        //開始計算 
        for(int i=0;i<roo;i++)
                for(int j=0;j<col;j++)
                    if(grid[i][j]<mmin[i]) mmin[i]=grid[i][j];
            for(int i=0;i<roo;i++){
                ans=ans+mmin[i];
                for(int j=0;j<col;j++){
                        grid[i][j]=grid[i][j]-mmin[i];
                        sum[i]=sum[i]+grid[i][j];
                    }
            }
            for(int i=1;i<roo;i++)          
                if(sum[i]!=sum[i-1]){ 
                    ans=-1;break;           
            }
            if(ans==-1){
                cout<<"-1"<<endl;
                continue;
            }            
            for(int j=0;j<col;j++){
                if(grid[0][j]!=0){
                    ans=ans+grid[0][j];
                }
            }
            cout<<ans<<endl;
            for(int i=0;i<roo;i++){
                for(int j=0;j<mmin[i];j++)
                det?cout<<"col "<<i+1<<endl:cout<<"row "<<i+1<<endl; //判斷是否轉置過 控制相應輸出 注意這裡是i                    
            }
            for(int j=0;j<col;j++){
                if(grid[0][j]!=0){              
                    for(int i=0;i<grid[0][j];i++){
                        det?cout<<"row "<<j+1<<endl:cout<<"col "<<j+1<<endl;//判斷是否轉置過 控制相應輸出 注意這裡是j
                    }
                }
            }       
    }
    return 0;
}