Description

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i)      empty the pot i to the drain;
3. POUR(i,j)    pour from pot i to pot j; after this operation either the pot j
   is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input

On the first and only line are the numbers A

Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible

Sample Input

3 5 4

Sample Output

6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

題意：有兩個空瓶 a，b是它們的容量，c是容量目標， 可以有三種操作 充滿任意一瓶，倒空任意一瓶，將任意一瓶倒入另一瓶（能剩下但不能溢位）；求任意一瓶的體積
達到目標體積所需要的最小運算元，並依此輸出該操作。
程式碼：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int m,n,k;
int vis[105][105];
char opr[20][20]= {" " , "FILL(1)" , "FILL(2)" , "DROP(1)" , "DROP(2)" , "POUR(1,2)" , "POUR(2,1)" };     //共6種操作
struct node
{
    int x,y,step;
    int w[200];                    //用來記錄路徑的陣列陣列  
};
void bfs( )
{
    int i,j;
    int kx,ky;
    memset(vis,0,sizeof(vis));
    queue<node>q;
    node now,next;
    now.x=0,now.y=0,now.step=0;
    q.push(now);
    vis[0][0]=1;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(now.x==k||now.y==k)
        {
            cout<<now.step<<endl;
            for( i=1; i<=now.step; i++)
            {
                cout<<opr[now.w[i]]<<endl;
            }
            return;
        }
        for(i=0; i<6; i++)        // 共六種操作
        {
            if(i==0)
            {
                next.y=now.y;
                next.x=m;
                now.w[now.step+1]=1;           //隨時更新
            }
            else if(i==1)
            {
                next.x=now.x;
                next.y=n;
                now.w[now.step+1]=2;
            }
            else if(i==2)
            {
                next.y=now.y;
                next.x=0;
                now.w[now.step+1]=3;
            }
            else if(i==3)
            {
                next.x=now.x;
                next.y=0;
                now.w[now.step+1]=4;
            }
            else if(i==4)
            {
                if(n-now.y>now.x)          //每種pour 應有兩種情況
                {
                    next.x=0;
                    next.y=now.x+now.y;
                }
                else
                {
                    next.y=n;
                    next.x=now.x-n+now.y;
                }
                now.w[now.step+1]=5;
            }
            else if(i==5)
            {
                if(m-now.x>now.y)
                {
                    next.y=0;
                    next.x=now.x+now.y;
                }
                else
                {
                    next.x=m;
                    next.y=now.y-m+now.x;
                }
                now.w[now.step+1]=6;             
            }
            if(vis[next.x][next.y]==1)
                continue;
            vis[next.x][next.y]=1;
            next.step=now.step+1;
            for(j=1; j<=next.step; j++)            //記錄之前的行動  
            {
                next.w[j]=now.w[j];
            }
            q.push(next);
        }
    }
    cout<<"impossible"<<endl;             //不要忘了這個情況
    return;
}
int main()
{
    int i;
    while(cin>>m>>n>>k)
    {
        bfs();
    }
    return 0;
}