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 math
import sys
from bisect import bisect_right, bisect_left, insort_right
from collections import Counter, defaultdict
from heapq import heappop, heappush
from itertools import accumulate, permutations, combinations
from sys import stdout

R = lambda: map(int, input().split())
n, m, x, y = R()
cs = [[0] * (m + 1), [0] * (m + 1)]
for _ in range(n):
    for i, c in enumerate(input()):
        cs[0][i] += c == '#'
        cs[1][i] += c == '.'
dp = [([math.inf] * (m + 1) + [0]) for i in range(2)]
dp[0][x - 1] = sum(cs[0][i] for i in range(x))
dp[1][x - 1] = sum(cs[1][i] for i in range(x))
for i in range(x, m):
    for t in range(2):
        acc = sum(cs[t][j] for j in range(i, i - x, -1))
        for j in range(i - x, max(-2, i - y - 1), -1):
            dp[t][i] = min(dp[t][i], acc + dp[1 - t][j])
            acc += cs[t][j - 1]
print(min(dp[0][m - 1], dp[1][m - 1]))