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 n;
char mp[5][1005];
int a[5], b[1005];
int dp[1005][1 << 12];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%I64d", &a[i]);
  for (int i = 1; i <= 4; i++) scanf("%s", mp[i] + 1);
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      b[j] += (mp[i][j] == '*') << (i - 1);
    }
  }
  memset(dp, 0x3f, sizeof dp);
  dp[1][b[1] + (b[2] << 4) + (b[3] << 8)] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = (1 << 12) - 1; j >= 0; j--) {
      if (dp[i][j] == 0x3f3f3f3f) continue;
      if ((j & 0xf) == 0) {
        int t = (j >> 4) + (b[i + 3] << 8);
        dp[i + 1][t] = min(dp[i + 1][t], dp[i][j]);
      }
      for (int k = 1; k < 4; k++) {
        for (int l = 0; l + k <= 4; l++) {
          int t = 0;
          for (int kk = 1; kk <= k; kk++) {
            t = t << 4;
            for (int ll = l; ll < l + k; ll++) {
              t += (1 << ll);
            }
          }
          t = j & ~t;
          dp[i][t] = min(dp[i][t], dp[i][j] + a[k]);
        }
      }
      dp[i + 1][0] = min(dp[i + 1][0], dp[i][j] + a[4]);
    }
  }
  printf("%d\n", dp[n][0]);
  return 0;
}
