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 "#".
def minv(DP, j, x):
    wyn = 10 ** 7
    for i in range(x, len(DP)):
        wyn = min(wyn, DP[i][j])
    return wyn

def main():
    n, m, x, y = map(int,input().split())
    tab = [0] * m

    for i in range(n):
        tmp = input()
        for j in range(m):
            if tmp[j] == "#":
                tab[j] += 1
    
    DP_W = [[10 ** 7] * m for i in range(y)]
    DP_B = [[10 ** 7] * m for i in range(y)]
    DP_W[-1][0] = tab[0]
    DP_B[-1][0] = n - tab[0]

    for i in range(1, m):
        DP_W[-1][i] = minv(DP_B, i - 1, x - 1) + tab[i]

        for j in range(y - 1):
            DP_W[j][i] = DP_W[j + 1][i - 1] + tab[i]

        DP_B[-1][i] = minv(DP_W, i - 1, x - 1) + (n - tab[i])

        for j in range(y - 1):
            DP_B[j][i] = DP_B[j + 1][i - 1] + (n - tab[i])

    print(min(minv(DP_W, -1, x - 1), minv(DP_B, -1, x - 1)))

main()

