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 = []
for i in range(n):
	s.append(input())
dp = [[[1e18,1e18] for i in range(y+1)] for j in range(m)]
ct = [0 for i in range(m)]
for i in range(n):
	for j in range(m):
		if(s[i][j] == '.'): 
			ct[j] += 1
dp[0][1][1] = ct[0]
dp[0][1][0] = n-ct[0]
for i in range(1,m):
	for j in range(y+1):
		dp[i][j][1] = dp[i-1][j-1][1]+ct[i]
		dp[i][j][0] = dp[i-1][j-1][0]+n-ct[i]
		if(j >= x):
			dp[i][1][0] = min(dp[i][1][0],dp[i-1][j][1]+n-ct[i])
			dp[i][1][1] = min(dp[i][1][1],dp[i-1][j][0]+ct[i])
ans = min(min(dp[-1]))
print(ans)