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 "#".
from math import inf

n,m,x,y=map(int,input().split())
matx=[]
for i in range(n):
    matx.append(list(input().strip()))

d=[[0,0]]
for i in range(m):
    c=0
    for j in range(n):
        if matx[j][i]=='.':
            c+=1
    d.append([c,n-c])

print(d)
for i in range(1,m+1):
    d[i][0]+=d[i-1][0]
    d[i][1]+=d[i-1][1]
#dp[n][0,1][p]

dp=[[1000,1000]for j in range(m+1)]
dp[0]=[0,0]
dp[x]=[d[x][0],d[x][1]]
for i in range(x,m+1):
    for j in range(x,min(y+1,i+1)):
        dp[i][0]=min(dp[i-j][1]+d[i][0]-d[i-j][0],dp[i][0])

        dp[i][1]=min(dp[i-j][0]+d[i][1]-d[i-j][1],dp[i][1])
#A&print(dp)
print(min(dp[-1]))
