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())
cols = []
for i in range(n):
  line = input()
  for j in range(m):
    if i ==0:
      cols.append(0)
    if line[j]=='.':
      cols[j]+=1

def valid(j):
  return  -1<j<m

mem = {}
c =0
def paint(j=0,prev=2,width = 0):
  global c
  c+=1
  if not valid(j):
    if x<= width <=y:
      return 0
    return inf
  if mem.get((j,width,prev),-1) != -1:
    return mem[j][width][prev] 

  if prev == 0:
    if width <x:
      mem[(j,width,prev)] = n- cols[j] + paint(j+1,0,width+1)
      return mem[(j,width,prev)] 
    if width == y :
      mem[(j,width,prev)] = cols[j] + paint(j+1,1,1)
      return mem[(j,width,prev)] 
  if prev == 1: 
    if width <x:
      mem[(j,width,prev)] = cols[j]+ paint(j+1,1,width+1)
      return mem[(j,width,prev)] 
    if width == y :
      mem[(j,width,prev)] = n- cols[j] + paint(j+1,0,1)
      return mem[(j,width,prev)] 

  p1 = n-cols[j]+ paint(j+1,0, width+1 if prev == 0 else 1)
  p2 = cols[j] + paint(j+1,1,width+1 if prev==1 else 1)
  mem[(j,width,prev)] = min(p1,p2)
  return mem[(j,width,prev)]




print(paint())


# print('in',c)