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 "#".
from math import inf
n,m,x,y = map(int, input().split())
A = []
for _ in range(n):
    A.append(input().strip())

white_cost = []
black_cost = []
for i in range(m):
    white_cost.append(sum(1 if A[j][i] == "#" else 0 for j in range(n)))
    black_cost.append(n-white_cost[-1])
cost = [white_cost, black_cost]

# dp = (best at index i, with j painted black before i)
dp = [[[inf, inf]] * (m) for _ in range(m)]
dp[0][0][0] = cost[0][0] 
dp[0][0][1] = cost[1][0]

for i in range(1,m):
    for w in range(1,i+1):
        for c in range(2):
            if w > 0 and w < x:
                dp[i][w][c] = dp[i-1][w-1][c] + cost[c][i]
            elif w > y:
                dp[i][w][c] = dp[i-1][1][1 - c] + cost[1-c][i]
            else:
                dp[i][w][c] = min(
                        dp[i-1][w-1][c] + cost[c][i],
                        dp[i-1][1][1 - c] + cost[1 - c][i]
                            )


print(min(v for arr in dp[-1] for v in arr))
