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())
l=[]
cnt =[]
for i in range(n):
    l1=input()
    l.append(l1)
for i in range(m):
    c=0
    for j in range(n):
        if l[j][i]=='.':
            c+=1 
    cnt.append(c)
pre =[0]
for i in range(m):
    pre.append(pre[-1]+cnt[i])
dp=[[0]*(m+1) for j in range(m+1)]
for  i in range(1,m+1):
    dp[0][i]=10**18
    dp[1][i]=10**18
    
    
    for j in range(x,y+1):
        if i>=j: 
            dot=pre[i]-pre[i-j]
            has=n*j-dot 
            dp[0][i]=min(dp[0][i],dp[1][i-j]+dot)
            dp[1][i]=min(dp[1][i],dp[0][i-j]+has)
print(min(dp[0][m],dp[1][m]))
            
            