You've got an *n* × *m* pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.

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*.
The first line contains four space-separated integers *n*, *m*, *x* 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 "#".
#!/usr/bin/env python3
# created : 2020. 12. 31. 23:59

import os
from sys import stdin, stdout


def solve(tc):
    n, m, x, y = map(int, stdin.readline().split())

    grid = []
    for i in range(n):
        grid.append(list(stdin.readline().strip()))

    dp = [[[n*m, n*m] for j in range(y+1)] for i in range(m+1)]
    dp[0][0][0] = dp[0][0][1] = 0

    for i in range(m):
        blacks = 0
        for j in range(n):
            if grid[j][i] == '#':
                blacks += 1

        for k in range(x):
            dp[i+1][k+1][0] = min(dp[i+1][k+1][0], dp[i][k][0] + n - blacks)
            dp[i+1][k+1][1] = min(dp[i+1][k+1][1], dp[i][k][1] + blacks)

        for k in range(x, y):
            dp[i+1][k+1][0] = min(dp[i+1][k+1][0], dp[i][k][0] + n - blacks)
            dp[i+1][k+1][1] = min(dp[i+1][k+1][1], dp[i][k][1] + blacks)
            dp[i+1][1][0] = min(dp[i+1][1][0], dp[i][k][1] + n - blacks)
            dp[i+1][1][1] = min(dp[i+1][1][1], dp[i][k][0] + blacks)

        dp[i+1][1][0] = min(dp[i+1][1][0], dp[i][y][1] + n - blacks)
        dp[i+1][1][1] = min(dp[i+1][1][1], dp[i][y][0] + blacks)

    ans = n*m
    for k in range(x, y+1):
        ans = min(ans, dp[m][k][0])
        ans = min(ans, dp[m][k][1])
    print(ans)


tcs = 1
# tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
    solve(tc)
    tc += 1
