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 "#".
inf = 10**10

def main():
    n, m, x, y = list(map(int, input().split()))
    table = [[1 if i == "#" else 0 for i in input()]for j in range(n)]
    cnt = [0] * m
    for i in range(n):
        for j in range(m):
            cnt[j] += table[i][j]
    dp1 = [[inf for j in range(0, y+1)] for i in range(m)]
    dp2 = [[inf for j in range(0, y+1)] for i in range(m)]
    dp1[0][1], dp2[0][1] = cnt[0], n - cnt[0]
    for i in range(1, m):
        for j in range(1, y+1):
            if j <= i:
                if j == 1:
                    dp1[i][1] = min(dp1[i][j], min(dp2[i-1]) + cnt[i])
                    dp2[i][1] = min(dp2[i][j], min(dp1[i-1]) + n - cnt[i])
                else:
                    dp1[i][j] = min(dp1[i][j], dp1[i-1][j-1] + cnt[i])
                    dp2[i][j] = min(dp2[i][j], dp2[i-1][j-1] + n - cnt[i])
    print(min(min(dp1[m-1][x:y+1]), min(dp2[m-1][x:y+1])))




if __name__ == "__main__":
    main()