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())
a=[]
for i in range(n):
    a.append(list(input()))

dp=[[-10000000,-10000000] for i in range(m+1)]
dp[0]=[0,0]
cost=[0 for i in range(m+1)]
for i in range(m):
    cost[i+1]=cost[i]
    for j in range(n):
        if a[j][i]==".":
            cost[i+1]+=1

for i in range(1,m+1):
    if i<x:
        dp[i][0]=-1
        dp[i][1]=-1
        continue
    for nxt in range(i-x,max(-1,i-y-1),-1):
        dp[i][0]=max(dp[i][0],dp[nxt][1]+cost[i]-cost[nxt])
        dp[i][1]=max(dp[i][1],dp[nxt][0]+(n*(i-nxt)-(cost[i]-cost[nxt])))
print(n*m-max(dp[m][0],dp[m][1]))