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 cost[5];
char buff[1005];
string arr[5];
int n = 4, m;
int DP[4][1001][1 << 4][1 << 4][1 << 4];
int mask[5][5][5];
int solve(int ind, int col, int m1, int m2, int m3) {
  if (ind && (m1 & (1 << (ind - 1)))) m1 -= 1 << (ind - 1);
  if (col == m) return 0;
  if (ind == n) return solve(0, col + 1, m2, m3, 0);
  int &temp = DP[ind][col][m1][m2][m3];
  if (temp != -1) return temp;
  temp = 2147483647;
  if ((m1 & (1 << ind)) || arr[ind][col] == '.') {
    temp = solve(ind + 1, col, m1, m2, m3);
  }
  for (int i = 1; i <= 3; i++) {
    if (ind + i > n || col + i > m) break;
    temp = min(temp,
               cost[i - 1] + solve(ind + 1, col, m1 | mask[ind][i][0],
                                   m2 | mask[ind][i][1], m3 | mask[ind][i][2]));
  }
  if (ind == 0 && col + 4 <= m) {
    temp = min(temp, cost[3] + solve(0, col + 1, mask[0][4][0], mask[0][4][1],
                                     mask[0][4][2]));
  }
  return temp;
}
int main() {
  cin >> m;
  for (int i = 0; i < n; i++) scanf("%d", &cost[i]);
  for (int i = 0; i < n; i++) {
    scanf("%s", buff);
    arr[i] = string(buff);
  }
  for (int i = 1; i <= 4; i++) {
    int tmp = (1 << i) - 1;
    for (int j = 0; j < 4; j++) {
      if (j + i > 4) break;
      for (int k = 0; k < i; k++) mask[j][i][k] = tmp;
      tmp *= 2;
    }
  }
  memset(DP, -1, sizeof(DP));
  cout << solve(0, 0, 0, 0, 0);
  return 0;
}
