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

n, m, x, y = sys.stdin.readline().strip().split(' ')
n, m, x, y = int(n), int(m), int(x), int(y)

a = [0 for i in range(m)]

for i in range(n):
    str = sys.stdin.readline()
    for j in range(m):
        if str[j] == '.':
            a[j] = a[j] + 1
            
pref = [[0 for i in range(m+1)] for c in {0, 1}]
pref[0][0] = a[0]
pref[1][0] = n - a[0]

for i in range(1, m):
    pref[0][i] = a[i] + pref[0][i-1]
    pref[1][i] = n - a[i] + pref[1][i-1]


inf = n*m + 1000;   

#f[c][i]
f = [[inf for j in range(m+1)] for c in {0, 1}]
f[0][0] = 0
f[1][0] = 0

def dp(c, i):
    if i < 0:
        return inf
    return f[c][i]
    
def diff(c, frm, to): #[frm; to]
    if frm == 0:
        return pref[c][to]
    return pref[c][to] - pref[c][frm-1]


for i in range(1, m + 1):
    for c in {0, 1}:
        f[c][i] = min([dp(1-c, i-j) + diff(c, i - j, i-1) for j in range(x, y+1)])

#print(f)       
#print(pref)
print (min(f[0][m], f[1][m]))