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
from math import inf
n,m,x,y = map(int, input().split())
A = []
for _ in range(n):
    A.append(input().strip())

num_black = []
for i in range(m):
    num_black.append(sum(1 if A[j][i] == "#" else 0 for j in range(n)))

# (col number, how many cols before I've colored)
#jdp = [[inf] * y for _ in range(m)]
#jdp[0][0] = 

@lru_cache(maxsize=None)
def opt(i, colored):
    if i >= m:
        return 0
    if colored > 0 and colored < x:
        return opt(i+1, colored+1) + n - num_black[i]
    if colored == y:
        return opt(i+1, 0) + num_black[i]
    return min(
            opt(i+1, 0) + num_black[i],
            opt(i+1, colored+1) + n - num_black[i]
            )

print(opt(0,0))
