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 "#".
n,m,x,y = map(int,input().split())
s = [input() for _ in range(n)]
memo = [[[-1, -1] for i in range(y+1)] for j in range(m)]
ct = [0 for i in range(m)]

def dp(i, cnt, f):
	if i == m:
		return 0 if cnt >= x else 1e18
	if memo[i][cnt][f] != -1: 
	    return memo[i][cnt][f]
	res = 1e18
	val = n - ct[i] if f == 0 else ct[i]
	if cnt >= x:
		res = min(res, dp(i+1, 1, 1-f) + n - val)
	if cnt < y:
		res = min(res, dp(i+1, cnt+1, f) + val)
	memo[i][cnt][f] = res
	return memo[i][cnt][f]
 
for i in range(n):
	for j in range(m):
		if(s[i][j] == '.'): 
			ct[j] += 1
print(min(dp(0, 0, 0), dp(0, 0, 1)))
