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 "#".
from functools import lru_cache
n, m ,x, y = map(int, input().split())
s = [{'#' : 0, ".": 0 } for i in range(m)]

@lru_cache(maxsize=None)
def f(mn, mx, i, last=None):
    global x, y

    if i == -1:
        return 0

    if mn != 0 and last!=None:
        return f(mn-1, mx-1, i-1, last) + s[i][last]
    elif mx == 0:
        if last == '.':
            return f(x, y, i-1, '#') + s[i]['#']
        else:
            return f(x, y, i - 1, '.') + s[i]['.']
    else:
        if last == '.':
            return min(f(mn - 1, mx - 1, i - 1, '.') + s[i]['.'], f(x -1, y-1, i - 1, '#') + s[i]['#'])
        else:
            return min(f(x-1,y-1, i - 1, '.') + s[i]['.'], f(mn-1, mx-1, i - 1, '#') + s[i]['#'])



for i in range(n):
    ls = input()
    for j in range(len(ls)):
        s[j][ls[j]]+=1



print(f(x,y,len(s)-1, None))