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 N = 1005;
int n, tmp, a[5][5], num[5][5], c[5], cnt[N], f[N][1 << 16];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= n; j++) {
      char c;
      cin >> c;
      cnt[j] <<= 1;
      if (c == '*') cnt[j] |= 1;
    }
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= 4 - i + 1; j++) {
      for (int k = 1; k <= 4; k++)
        for (int l = 1; l <= 4; l++) a[k][l] = 1;
      for (int k = 1; k <= j; k++)
        for (int l = 0; l < j; l++) a[k][i + l] = 0;
      int sta = 0;
      for (int k = 4; k >= 1; k--)
        for (int l = 1; l <= 4; l++) sta = (sta << 1) + a[k][l];
      num[i][j] = sta;
    }
  for (int i = 4; i >= 1; i--) {
    tmp <<= 4;
    tmp += cnt[i];
  }
  memset(f, 0x3f, sizeof(f));
  f[1][tmp] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = (1 << 16) - 1; j >= 0; j--) {
      if (f[i][j] == 0x3f3f3f3f) continue;
      if ((j & 15) == 0) {
        int sta = (cnt[i + 4] << 12) | (j >> 4);
        f[i + 1][sta] = min(f[i + 1][sta], f[i][j]);
      }
      for (int k = 1; k <= 4; k++)
        for (int l = 1; l <= 4 - k + 1; l++)
          f[i][j & num[k][l]] = min(f[i][j & num[k][l]], f[i][j] + c[l]);
    }
  printf("%d", f[n][0]);
  return 0;
}
