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 "#".
n, m, x, y = map(int, input().split())
u = [i.count('.') for i in list(zip(*[input() for i in range(n)]))]
v = [n - i for i in u]
a, b, s = [u[0]], [v[0]], x - 1
for i in range(1, x):
    a, b = [1000001] + [j + u[i] for j in a], [1000001] + [j + v[i] for j in b]
for i in range(x, min(y, m)):
    a, b = [min(b[s: ]) + u[i]] + [j + u[i] for j in a], [min(a[s: ]) + v[i]] + [j + v[i] for j in b]
for i in range(min(y, m), m):
    a, b = [min(b[s: ]) + u[i]] + [j + u[i] for j in a[: -1]], [min(a[s: ]) + v[i]] + [j + v[i] for j in b[: -1]]
print(min(min(a[s: ]), min(b[s: ])))