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 process(M, x, y):
    curr = {('#', 0): 0, ('.', 0): 0}
    n = len(M)
    m = len(M[0])
    for j in range(m):
        curr2 = {}
        counts = {'#': 0, '.': 0}
        for i in range(n):
            counts[M[i][j]]+=1
        cost1 = counts['.']
        cost2 = counts['#']
        for X, Y in curr:
            if X=='#' and (Y+1 <= y):
                x2 = X
                y2 = Y+1
                if (x2, y2) not in curr2:
                    curr2[(x2, y2)] = float('inf')
                curr2[(x2, y2)] = min(cost1+curr[(X, Y)], curr2[(x2, y2)])
            if X=='#' and (x <= Y <= y):
                x2 = '.'
                y2 = 1
                if (x2, y2) not in curr2:
                    curr2[(x2, y2)] = float('inf')
                curr2[(x2, y2)] = min(cost2+curr[(X, Y)], curr2[(x2, y2)])
            if X=='.' and (Y+1 <= y):
                x2 = X
                y2 = Y+1
                if (x2, y2) not in curr2:
                    curr2[(x2, y2)] = float('inf')
                curr2[(x2, y2)] = min(cost2+curr[(X, Y)], curr2[(x2, y2)])
            if X=='.' and (x <= Y <= y):
                x2 = '#'
                y2 = 1
                if (x2, y2) not in curr2:
                    curr2[(x2, y2)] = float('inf')
                curr2[(x2, y2)] = min(cost1+curr[(X, Y)], curr2[(x2, y2)])
        curr = curr2
    return min(curr.values())

n, m, x, y = [int(x) for x in input().split()]
M = []
for i in range(n):
    row = input()
    M.append(row)
print(process(M, x, y))