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())
grid=[]
for _ in range(n):
    grid.append(list(input()))
col = []
for i in range(m):
    white=0
    for ii in range(n):
        white += (grid[ii][i] == ".")
    col.append([white,n-white])
for i in range(1,len(col)):
    col[i]=[col[i-1][0]+col[i][0],col[i-1][1]+col[i][1]]
col.append([0,0])
# [[],[white,black],[],[]] optimal val
dp=[[1e18,1e18] for _ in range(m+1)];dp[-1]=[0,0]
for i in range(x-1):
    dp[i]=[0,0]
for i in range(x-1,m):
    for ii in range(x-1,y+1):
        if i-ii<-1:continue
        dp[i][0]=min(dp[i][0],dp[i-ii][1]+(col[i][1]-col[i-ii][1]))
        dp[i][1]=min(dp[i][1],dp[i-ii][0]+(col[i][0]-col[i-ii][0]))
print(min(dp[-2]))