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 = [0]*m
for i in range(n):
    a = list(input())
    for j in range(m):
        s[j] += 1 if a[j] == '#' else 0

ss = [*s]
for i in range(m -2, -1, -1):
    ss[i] = ss[i] + ss[i+1]

dp = [[0,0] for i in range(m)]
for i in range(m):
    a_0 = []
    a_1 = []
    for z in range(x,y+1):
        if i - z >= -1:
            c1 = ss[i-z + 1]
            if i < m-1:
                c1 -= ss[i+1]
            a_0.append(c1)
            a_1.append(n*z - c1)
            if i - z >= 0:
                a_0[len(a_0) -1] += dp[i-z][1]
                a_1[len(a_1) -1] += dp[i-z][0]
        else:
            break
    dp[i][0] = min(a_0)
    dp[i][1] = min(a_1)
print(min(dp[m-1]))