Description



You've got an n × m

A picture is a barcode if the following conditions are fulfilled:

  • All pixels in each column are of the same color.
  • The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.


Input



The first line contains four space-separated integers nmx and y (1 ≤ n, m, x, y ≤ 1000; x ≤ y).

Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".


Output



In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.


Sample Input



Input



6 5 1 2 ##.#. .###. ###.. #...# .##.# ###..



Output



11



Input



2 5 1 1 ##### .....



Output



5


Hint



In the first test sample the picture after changing some colors can looks as follows:



.##.. .##.. .##.. .##.. .##.. .##..



In the second test sample the picture after changing some colors can looks as follows:



#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 2005;
int f[maxn][2];
int x, y, n, m, i, j, a[maxn],b[maxn];
char s[maxn];

int main(){
while (cin >> n >> m >> x >> y)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(f, 1, sizeof(f));
for (i = 1; i <= n; i++)
{
scanf("%s", s);
for (j = 0; j < m;j++)
if (s[j] == '#') a[j + 1]++; else b[j + 1]++;
}
for (j = 1; j <= m; j++) a[j + 1] += a[j], b[j + 1] += b[j];
f[0][0] = f[0][1] = 0;
for (i = 0; i < m; i++)
{
for (j = x; j <= y; j++)
{
f[i + j][0] = min(f[i + j][0], f[i][1] + a[i + j] - a[i]);
f[i + j][1] = min(f[i + j][1], f[i][0] + b[i + j] - b[i]);
}
}
cout << min(f[m][0], f[m][1]) << endl;
}
return 0;
}