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 "#".
import sys
from functools import lru_cache
sys.setrecursionlimit(1 << 30)

def solve():
    n,m,x,y = list(map(int, input().split())) 
    grid = [] 
    for _ in range(n):
        grid.append(list(input())) 
    rows = [] 
    for j in range(m):
        counts = 0
        for i in range(n):
            if grid[i][j] == '#':
                counts += 1 
        rows.append((counts,n- counts )) 
    # print(rows)
    @lru_cache 
    def helper(i,prev,count):
        if i == m:
            return 0 if x <= count <= y else float('inf') 
        cont = float('inf') 
        if count + 1 <= y:
            cont = helper(i + 1, prev, count + 1) + rows[i][1 if prev == -1 else 0] 
        return min(cont, helper(i + 1,-1*prev,1) + rows[i][0 if prev == -1 else 1]) 

    return min(helper(0,-1,1), helper(0,1,1)) 

print(solve())