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 collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect

#?############################################################


def isPrime(x):
    if(x == 1):
        return False
    for i in range(2, x):
        if i*i > x:
            break
        if (x % i == 0):
            return False
    return True

#?############################################################


def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den, p - 2, p)) % p


#?############################################################

def primeFactors(n):
    l = []
    while n % 2 == 0:
        l.append(2)
        n = n / 2
    for i in range(3, int(math.sqrt(n))+1, 2):
        while n % i == 0:
            l.append(int(i))
            n = n / i
    if n > 2:
        l.append(n)
    return list(set(l))


#?############################################################

def power(x, y, p):
    res = 1
    x = x % p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res

#?############################################################


def sieve(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime


#?############################################################

def digits(n):
    c = 0
    while (n > 0):
        n //= 10
        c += 1
    return c

#?############################################################


def ceil(n, x):
    if (n % x == 0):
        return n//x
    return n//x+1

#?############################################################


def mapin():
    return map(int, input().split())

#?############################################################


input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>opgf
n, m, x, y = mapin()
l = [[0, 0] for i in range(m)]
for i in range(n):
    dd = input()
    l.append([0, 0])
    for j in range(m):
        if(dd[j] == "."):
            l[j][0] += 1
        else:
            l[j][1] += 1


dp = [[0, 0] for i in range(m+1)]
pb, pw = [0], [0]
for i in range(m):
    pb.append(pb[-1]+l[i][0])
    pw.append(pw[-1]+l[i][1])
    

dp = [[float('inf')]*(m+1) for i in range(2)]
dp[0][0] = dp[1][0] = 0
for i in range(1, m+1):
    for j in range(x, y+1):
        if i-j >= 0:
            dp[0][i] = min(dp[0][i], dp[1][i-j]+pb[i]-pb[i-j])
            dp[1][i] = min(dp[1][i], dp[0][i-j]+pw[i]-pw[i-j])
print(min(dp[0][-1], dp[1][-1]))
