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 sys import stdin,stdout , setrecursionlimit
from threading import Thread , stack_size
setrecursionlimit(2**31-1)
def print(*args, end='\n', sep=' ') -> None:
    stdout.write(sep.join(map(str, args)) + end)

def dp(col,wid,last):
    if col==m:
        if wid<mini or wid>maxi:
            return 100000000
        return 0

    if (col,wid,last) in memo:
        return memo[(col,wid,last)]
    dots=100000000
    if col==0 or (last==0 and wid+1<=maxi) or (last==1 and wid>=mini):
        if last==1 : x=1
        else: x=wid+1
        dots=n-dotcou[col]+dp(col+1,x,0)
    black=100000000
    if col==0 or (last==1 and wid<maxi) or (last==0 and wid>=mini):
        if last==0 : x=1
        else: x=wid+1
        black=dotcou[col]+dp(col+1,x,1)
    ans=min(black,dots)
    memo[(col,wid,last)]=ans
    return ans
    

def Solve():
    global n,m,mini,maxi,dotcou,memo
    n,m,mini,maxi=map(int,input().split())  ; dotcou=[0]*m ; memo={}
    for i in range(n):
        s=input().strip()
        for j in range(m):
            if s[j]==".":dotcou[j]+=1
    print(dp(0,0,2))

stack_size(10**8)
Thread(target=Solve).start()