You are given a matrix f with 4 rows and n columns. Each element of the matrix is either an asterisk (*) or a dot (.).

You may perform the following operation arbitrary number of times: choose a square submatrix of f with size k × k (where 1 ≤ k ≤ 4) and replace each element of the chosen submatrix with a dot. Choosing a submatrix of size k × k costs ak coins.

What is the minimum number of coins you have to pay to replace all asterisks with dots?

Input

The first line contains one integer n (4 ≤ n ≤ 1000) — the number of columns in f.

The second line contains 4 integers a1, a2, a3, a4 (1 ≤ ai ≤ 1000) — the cost to replace the square submatrix of size 1 × 1, 2 × 2, 3 × 3 or 4 × 4, respectively.

Then four lines follow, each containing n characters and denoting a row of matrix f. Each character is either a dot or an asterisk.

Output

Print one integer — the minimum number of coins to replace all asterisks with dots.

Examples

Input

4
1 10 8 20
***.
***.
***.
...*


Output

9


Input

7
2 1 8 2
.***...
.***..*
.***...
....*..


Output

3


Input

4
10 10 1 10
***.
*..*
*..*
.***


Output

2

Note

In the first example you can spend 8 coins to replace the submatrix 3 × 3 in the top-left corner, and 1 coin to replace the 1 × 1 submatrix in the bottom-right corner.

In the second example the best option is to replace the 4 × 4 submatrix containing columns 2 – 5, and the 2 × 2 submatrix consisting of rows 2 – 3 and columns 6 – 7.

In the third example you can select submatrix 3 × 3 in the top-left corner and then submatrix 3 × 3 consisting of rows 2 – 4 and columns 2 – 4.
#include <bits/stdc++.h>
using namespace std;
int dp[1005][(1 << 4)][(1 << 4)][(1 << 4)][(1 << 4)];
int cost[(1 << 4)];
int mask[2005];
string str[4];
int n;
int calc(int idx, int col1, int col2, int col3, int col4) {
  if (idx >= n) return 0;
  if (dp[idx][col1][col2][col3][col4] != -1)
    return dp[idx][col1][col2][col3][col4];
  int ret = 123456789;
  if (col1 == 0)
    return dp[idx][col1][col2][col3][col4] =
               calc(idx + 1, col2, col3, col4, mask[idx + 4]);
  for (int i = 0; i < 4; i++) {
    if (col1 & (1 << i)) {
      int ncol1 = col1 ^ (1 << i);
      int val = cost[0] + calc(idx, ncol1, col2, col3, col4);
      ret = min(ret, val);
    }
  }
  for (int i = 0; i < 3; i++) {
    int f1 = col1 & (1 << i), f2 = col1 & (1 << (i + 1));
    int g1 = col2 & (1 << i), g2 = col2 & (1 << (i + 1));
    if (f1 + f2) {
      int ncol1 = col1;
      if (f1) ncol1 = ncol1 ^ (1 << i);
      if (f2) ncol1 = ncol1 ^ (1 << (i + 1));
      int ncol2 = col2;
      if (g1) ncol2 = ncol2 ^ (1 << i);
      if (g2) ncol2 = ncol2 ^ (1 << (i + 1));
      int val = cost[1] + calc(idx, ncol1, ncol2, col3, col4);
      ret = min(ret, val);
    }
  }
  for (int i = 0; i < 2; i++) {
    int f1 = col1 & (1 << i), f2 = col1 & (1 << (i + 1)),
        f3 = col1 & (1 << (i + 2));
    int g1 = col2 & (1 << i), g2 = col2 & (1 << (i + 1)),
        g3 = col2 & (1 << (i + 2));
    int h1 = col3 & (1 << i), h2 = col3 & (1 << (i + 1)),
        h3 = col3 & (1 << (i + 2));
    if (f1 + f2 + f3) {
      int ncol1 = col1, ncol2 = col2, ncol3 = col3;
      if (f1) ncol1 = ncol1 ^ (1 << i);
      if (f2) ncol1 = ncol1 ^ (1 << (i + 1));
      if (f3) ncol1 = ncol1 ^ (1 << (i + 2));
      if (g1) ncol2 = ncol2 ^ (1 << i);
      if (g2) ncol2 = ncol2 ^ (1 << (i + 1));
      if (g3) ncol2 = ncol2 ^ (1 << (i + 2));
      if (h1) ncol3 = ncol3 ^ (1 << i);
      if (h2) ncol3 = ncol3 ^ (1 << (i + 1));
      if (h3) ncol3 = ncol3 ^ (1 << (i + 2));
      int val = cost[2] + calc(idx, ncol1, ncol2, ncol3, col4);
      ret = min(ret, val);
    }
  }
  int val = cost[3] + calc(idx + 4, mask[idx + 4], mask[idx + 5], mask[idx + 6],
                           mask[idx + 7]);
  ret = min(ret, val);
  return dp[idx][col1][col2][col3][col4] = ret;
}
string gen(int num) {
  string str = "";
  while (num != 0) {
    int mod = num % 2;
    if (mod)
      str += '1';
    else
      str += '0';
    num = num / 2;
  }
  while ((int)str.size() != 4) str = "0" + str;
  reverse(str.begin(), str.end());
  return str;
}
int main() {
  scanf("%d", &n);
  memset(mask, 0, sizeof mask);
  for (int i = 0; i < 4; i++) scanf("%d", &cost[i]);
  for (int i = 0; i < 4; i++) cin >> str[i];
  for (int i = 0; i < n; i++) {
    mask[i] = 0;
    for (int j = 0; j < 4; j++) {
      if (str[j][i] == '*') {
        mask[i] = mask[i] + (1 << j);
      }
    }
  }
  memset(dp, -1, sizeof dp);
  int ans = calc(0, mask[0], mask[1], mask[2], mask[3]);
  cout << ans;
}
