Description

A robot is located at the top-left corner of a m×nm×n grid. The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid. There is a non-negative integer in each small grid which means the score that the robot get when it passes this grid. Now you need to calculate the top kk

Note: The top k total scores may contain same total score got from different paths.

Input

The first line of input are m and n,which denote the size of the grid. The second line of input is the kk. The next m lines are the score matrix. (1≤m,n,k≤1001≤m,n,k≤100,and 0≤score≤1000≤score≤100)

Output

There is only one line in the Output,which is a descending sequence consist of the top k total scores.

Sample1

Input

3 3
2
3 2 5
3 2 2
4 2 1

Output

13 13

Sample2

Input

4 3
3
7 1 5
4 6 2
4 2 1
5 7 3

Output

30 29 27

#include <iostream>
using namespace std;
const int maxn = 200;

int m, n, K;//矩陣m*n,輸出TOPK的值

int dp[maxn][maxn][maxn];
int a[maxn][maxn];

void solve()
{

	dp[1][1][1] = a[1][1];
	for (int k = 2; k <= K; k++) dp[1][1][k] = -1;

	for (int j = 2; j <= n; j++)
	{
		dp[1][j][1] = dp[1][j - 1][1] + a[1][j];
		for (int k = 2; k <= K; k++) dp[1][j][k] = -1;
	}

	for (int i = 2; i <= m; i++)
	{
		for (int k = 1; k <= K; k++)
		{
			if (~dp[i - 1][1][k])  dp[i][1][k] = a[i][1] + dp[i - 1][1][k];
			else          dp[i][1][k] = -1;
		}


		for (int j = 2; j <= n; j++)
		{
			int ple = 1, pri = 1;
			int nex = 1;
			while (nex <= K)
			{
				int & x = dp[i - 1][j][ple];
				int & y = dp[i][j - 1][pri];

				if (x == -1 && y == -1) { dp[i][j][nex++] = -1; continue; }

				if (x >= y) { dp[i][j][nex++] = x + a[i][j]; ple++; }
				else { dp[i][j][nex++] = y + a[i][j]; pri++; }
			}

		}

	}


}


int main()
{
	
		cin >> m >> n >> K;
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				cin >> a[i][j];
			}
		}	

		solve();

		for (int k = 1; k <= K; k++)
		{
			cout << dp[m][n][k] << (k == K ? '\n' : ' ');//控制輸出格式
		}
	 

	return 0;
}