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
from os import path
rd = lambda:stdin.readline().strip()
wr = stdout.write
if(path.exists('input.txt')):
    stdin = open("input.txt","r")
#------------------------------------
memo = [[[-1]*3]*1009]*1009
a,b,l,r = map(int,rd().split())
mylist = []
for _ in range(a):
	mylist.append(rd())
dots= list()
for i in range(b):
	counter = 0 
	for j in range(a):
		if mylist[j][i] == '.':
			counter+=1
	dots.append(counter)
def solve(cur,width,last):
	global l,r,a,dots,memo
	if cur == b:
		if width > r or width < l:
			return 1000000000
		return 0

	if memo[cur][width][last] !=-1:
		return memo[cur][width][last]

	dot = 1000000000
	if (last == 1 and (width >= l and width<= r)) or(last ==0 and(width< r)) or last == 2:
		dot = (a-dots[cur])+solve(cur+1,0 if last == 1 else width+1,0)
	hash =1000000000
	if (last == 0 and (width >= l and width<= r)) or(last ==1 and(width< r)) or last == 2:
		hash = dots[cur]+solve(cur+1,0 if last == 0 else width+1,1)
	memo[cur][width][last] = min(dot,hash)
	return memo[cur][width][last]

print(solve(0,0,2)+1)
