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 maxn = 1002;
int f[maxn][(1 << 16)];
char s[5][maxn];
int a[5];
int M[5][5];
int cal(int i, int j) {
  int ans = 0;
  int k = j;
  while (k--) {
    for (int jj = 0; jj < j; jj++) ans |= (1 << i + jj);
    i += 4;
  }
  return ans;
}
void init() {
  for (int i = 0; i < 4; i++) {
    for (int j = 1; j <= 4; j++) {
      M[i][j] = cal(i, j);
    }
  }
}
int main() {
  init();
  int n;
  cin >> n;
  for (int i = 1; i <= 4; i++) scanf("%d", a + i);
  for (int i = 1; i <= 4; i++) scanf("%s", s[i]);
  memset(f, 0x3f, sizeof(f));
  int mask = 0;
  for (int i = 0; i < 4; i++) {
    for (int j = 1; j <= 4; j++)
      if (s[j][i] == '.') mask |= (1 << (j - 1 + 4 * i));
  }
  f[1][mask] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 4; j++) {
      for (int k = 4 - j; k > 0; k--) {
        for (int m = (1 << 16) - 1; m >= 0; m--) {
          f[i][m | M[j][k]] = min(f[i][m | M[j][k]], f[i][m] + a[k]);
        }
      }
    }
    mask = 0;
    if (i + 3 < n)
      for (int j = 1; j <= 4; j++)
        if (s[j][i + 3] == '.') mask += (1 << j - 1);
    mask <<= 12;
    for (int t = (1 << 12) - 1; t >= 0; t--) {
      int m = (t << 4) + 15;
      f[i + 1][(m >> 4) + mask] = min(f[i + 1][(m >> 4) + mask], f[i][m]);
      assert((m >> 4) + mask < (1 << 16));
    }
  }
  printf("%d\n", f[n][15]);
}
