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 "#".
rows, cols, min_width, max_width = map(int, input().split())

dp = [0] * cols

for i in range(rows):
    for col, c in enumerate(input().strip()):
        if c == ".":
            dp[col] += 1


def solve(col, prev, width):
    if col == cols:
        if min_width <= width <= max_width:
            return 0
        return float("inf")

    dots = dashes = float("inf")
    # dots 0
    if (
        col == 0
        or (prev == 0 and width + 1 <= max_width)
        or (prev == 1 and width >= min_width)
    ):
        dots = rows - dp[col] + solve(col + 1, 0, width + 1 if prev == 0 else 1)

    # dashes
    if (
        col == 1
        or (prev == 1 and width + 1 <= max_width)
        or (prev == 0 and width >= min_width)
    ):
        dashes = dp[col] + solve(col + 1, 1, width + 1 if prev == 1 else 1)

    return min(dots, dashes)


print(solve(0, -1, 0))
