【CF edu 30 C. Strange Game On Matrix】

time limit per test 1 second

memory limit per test  256 megabytes

input standard input

output standard output

Ivan is playing a strange game.

He has a matrix a with n rows and m columns. Each element of the matrix is equal to either 0 or 1. Rows and columns are 1-indexed. Ivan can replace any number of ones in this matrix with zeroes. After that, his score in the game will be calculated as follows:

1. Initially Ivan's score is 0;
2. In each column, Ivan will find the topmost 1 (that is, if the current column is j, then he will find minimum i such that a_i, j = 1). If there are no 1's in the column, this column is skipped;
3. Ivan will look at the next min(k, n - i + 1) elements in this column (starting from the element he found) and count the number of 1's among these elements. This number will be added to his score.

Of course, Ivan wants to maximize his score in this strange game. Also he doesn't want to change many elements, so he will replace the minimum possible number of ones with zeroes. Help him to determine the maximum possible score he can get and the minimum possible number of replacements required to achieve that score.

Input

The first line contains three integer numbers n, m and k (1 ≤ k ≤ n ≤ 100, 1 ≤ m ≤ 100).

Then n lines follow, i-th of them contains m integer numbers — the elements of i-th row of matrix a. Each number is either 0 or 1.

Output

Print two numbers: the maximum possible score Ivan can get and the minimum number of replacements required to get this score.

Examples

input

4 3 2
0 1 0
1 0 1
0 1 0
1 1 1

output

4 1

input

3 2 1
1 0
0 1
0 0

output

2 0

Note

In the first example Ivan will replace the element a_1, 2.

【翻译】给出01矩阵，要求删除1的点(可以不删),使得得分最大。得分计算方式：得分为每列的得分和，每列的得分计算方式为：从上往下第一个为1的位置向下的k长度区间内1的个数即分数(包括这个位置本身)。输出最高分数以及达成这个分数的最小删除数。

题解：
      ①贪心。

      ②每一列用单调性维护取最值就是了。

#include<stdio.h>
#include<algorithm>
#define go(i,a,b) for(int i=a;i<=b;i++)
#define ro(i,a,b) for(int i=a;i>=b;i--)
using namespace std;const int N=;
int n,m,k,G[N][N],sum[N],score,ans;
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    go(i,,n)go(j,,m)scanf("%d",&G[i][j]);
    go(j,,m)
    {
        int p,val=;
        ro(t,n,)sum[t]=sum[t+]+G[t][j];
        ro(t,n,)if(G[t][j]&&sum[t]-sum[min(t+k,n+)]>=val)
        val=sum[t]-sum[min(t+k,n+)],p=t;ans+=sum[]-sum[p];score+=val;
    }
    printf("%d %d\n",score,ans);return ;
}//Paul_Guderian

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