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
sys.setrecursionlimit(100000000)

n, m, x, y = map(int, input().split())
a = [0] * m

for i in range(n):
    for j, s in enumerate(input()):
        if s == '#': a[j] += 1
# print(*a)

inf = float('inf')
store = dict()
def solve(i, pn, pc, cost):
    if i == m: return inf if pn < x else cost
    if pn > y: return inf
    index = (i, pn, pc)
    if index in store: return store[index]

    res = None
    if pc == '#':
        res = min(solve(i + 1, pn + 1, '#', cost + (n - a[i])),
            solve(i + 1, 1, '.', cost + a[i]))
    else:
        res = min(solve(i + 1, pn + 1, '#', cost + (n - a[i])),
            solve(i + 1, 1, '.', cost + a[i]))
    store[index] = res
    return res

print(min(solve(0, 0, '#', 0), solve(0, 0, '.', 0)))
