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 inf = 2e9;
int n, a[10], f[5][1010][4100], bit[5][5];
char s[5][1010];
void upd(int &x, int v) { x = min(x, v); }
int id(int i, int j) {
  int now = (i - 1) * 4 + (j - 1);
  return 1 << now;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= 4; i++) {
    scanf("%s", s[i] + 1);
  }
  for (int i = 1; i < 4; i++) {
    for (int j = 1; j + i - 1 <= 4; j++) {
      for (int x = 1; x <= i; x++) {
        for (int y = 0; y < i; y++) {
          bit[i][j] += id(x, y + j);
        }
      }
      bit[i][j] ^= ((1 << 12) - 1);
    }
  }
  for (int i = 0; i <= 4; i++) {
    for (int j = 0; j <= n; j++) {
      for (int mask = 0; mask < (1 << 12); mask++) {
        f[i][j][mask] = inf;
      }
    }
  }
  f[4][0][0] = 0;
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = 1; j <= 4; j++) {
      if (s[j][i] == '*') {
        now |= 1 << (j - 1);
      }
    }
    for (int mask = 0; mask < (1 << 8); mask++) {
      upd(f[0][i][(mask << 4) | now], f[4][i - 1][mask]);
    }
    for (int j = 1; j <= 4; j++) {
      for (int mask = 0; mask < (1 << 12); mask++) {
        int v = f[j - 1][i][mask];
        if (v >= inf) {
          continue;
        }
        upd(f[j][i][mask], v);
        for (int k = 1; k < 4 && k <= j; k++) {
          if (mask == 8) {
          }
          upd(f[j][i][mask & bit[k][j - k + 1]], v + a[k]);
        }
      }
    }
    for (int j = 1; j <= 4 && j <= i; j++) {
      upd(f[4][i][0], f[4][i - j][0] + a[4]);
    }
  }
  cout << f[4][n][0] << endl;
  return 0;
}
