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())
grid = []
for i in range(n):
  grid.append(list(input()))
# [print(row) for row in grid]

def valid(j):
  return  -1<j<m
c = 0
def paint(j=0,selected = '',count=0):
  global c
  c+=1
  if not valid(j):
    # print(selected)
    return 0
  # cost of painting white
  white_cost = 0
  black_cost = 0
  for i in range(n):
    if grid[i][j] == '.':
      black_cost += 1
    if grid[i][j] == '#':
      white_cost +=1
  if len(selected) == 0:
    #try white
    p1 = white_cost + paint(j+1,'.',1)
    #try black
    p2 = black_cost + paint(j+1,'#',1)
    return min(p1,p2)
  if (selected == '.' and count < x):
    return white_cost + paint( j+1,'.',count+1)
  if selected == '#' and count==y :
    return white_cost + paint(j+1,'.',1)
  if selected == '.' and count==y :
    return black_cost + paint(j+1,'#',1)
  if (selected == '#' and count < x):
    return black_cost + paint( j+1,'#',count+1)
  if selected == '.' :
    p1 = white_cost + paint(j+1,'.',count+1)
    p2 = black_cost + paint(j+1,'#',1)
  if selected == '#' :
    p1 = white_cost + paint(j+1,'.',1)
    p2 = black_cost + paint(j+1,'#',count+1)
  return min(p1,p2)







print(paint())
# print(c)