Codeforces Round #FF (Div. 2) D. DZY Loves Modification 优先队列

D. DZY Loves Modification

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations.

Each modification is one of the following:

1. Pick some row of the matrix and decrease each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
2. Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.

DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.

Input

The first line contains four space-separated integers n, m, k and p (1 ≤ n, m ≤ 103; 1 ≤ k ≤ 106; 1 ≤ p ≤ 100).

Then n lines follow. Each of them contains m integers representing aij (1 ≤ aij ≤ 103) — the elements of the current row of the matrix.

Output

Output a single integer — the maximum possible total pleasure value DZY could get.

Examples

input

2 2 2 2
1 3
2 4

output

11

input

2 2 5 2
1 3
2 4

output

11

Note

For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:

1 1
0 0

For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:

-3 -3
-2 -2

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<bitset>
#include<set>
#include<map>
#include<time.h>
using namespace std;
#define LL long long
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e3+10,M=1e6+10,inf=1e9+10;
const LL INF=1e18+10,mod=1e9+7;
const double eps=(1e-8),pi=(4*atan(1.0));

LL a[N][N];
priority_queue<LL>col,row;
LL r[M],c[M];
int main()
{
    int n,m,k,p;
    scanf("%d%d%d%d",&n,&m,&k,&p);
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
        scanf("%lld",&a[i][j]);
    for(int i=1;i<=n;i++)
    {
        int r=0;
        for(int j=1;j<=m;j++)
        r+=a[i][j];
        row.push(r);
    }
    for(int j=1;j<=m;j++)
    {
        int c=0;
        for(int i=1;i<=n;i++)
        c+=a[i][j];
        col.push(c);
    }
    for(int i=1;i<=k;i++)
    {
        LL x=row.top();
        row.pop();
        r[i]=r[i-1]+x;
        x-=m*p;
        row.push(x);
    }
    for(int i=1;i<=k;i++)
    {
        LL x=col.top();
        col.pop();
        c[i]=c[i-1]+x;
        x-=n*p;
        col.push(x);
    }
    LL ans=-INF;
    for(int i=0;i<=k;i++)
        ans=max(ans,r[i]+c[k-i]-1LL*p*i*(k-i));
    printf("%lld\n",ans);
    return 0;
}