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())
a = [input() for i in range(n)]

c = [0] * m 
for j in range(m):
    for i in range(n):
        c[j]+=int(a[i][j] == '.')
pr = [0] * (m + 1)
for i in range(1, m + 1):
    pr[i] = pr[i - 1] + c[i - 1]

d = [[10 ** 10] * 2 for i in range(m + 1)]
d[0][0] = 0
d[0][1] = 0

for j in range(1, m + 1):
    for i in range(x, y + 1):
        if(j - i < 0):
            break # epic fail
        d[j][1] = min(d[j][1], d[j - i][0] + pr[j] - pr[j - i])
        d[j][0] = min(d[j][0], d[j - i][1] + (n * i - (pr[j] - pr[j - i])))
print(min(d[ m ][0], d[ m ][1]))

