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;
const int mx = 1 << 12;
int a[4], ms[4], dp[1007][mx], b[1007];
char c[4][1007];
int main() {
  int n, t, inf = 1e9 + 7;
  scanf("%d", &n);
  for (int i = 0; i < 4; i++) {
    scanf("%d", &a[i]);
    ms[i] = (1 << (i + 1)) - 1;
  }
  for (int i = 0; i < 4; i++) scanf("%s", &c[i]);
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 4; j++) b[i] += (c[j][i] == '*') << j;
    for (int j = 0; j < mx; j++) dp[i][j] = inf;
  }
  dp[0][b[0] + (b[1] << 4) + (b[2] << 8)] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = mx - 1; j >= 0; j--) {
      if (dp[i][j] == inf) continue;
      if (j % 16 == 0) {
        t = (j >> 4) + (b[i + 3] << 8);
        dp[i + 1][t] = min(dp[i + 1][t], dp[i][j]);
      }
      dp[i + 1][0] = min(dp[i + 1][0], dp[i][j] + a[3]);
      for (int k = 0; k < 3; k++) {
        for (int z = 0; z + k < 4; z++) {
          t = j;
          for (int p = 0; p <= k; p++) {
            t &= ~(ms[k] << (p * 4 + z));
          }
          dp[i][t] = min(dp[i][t], dp[i][j] + a[k]);
        }
      }
    }
  }
  printf("%d", dp[n][0]);
}
