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 b[1010], n, a1, a2, a3, a4, t;
const int inf = 0x3f3f3f3f;
int dp[1010][1 << 12], a[4],
    f[4] = {1, (1 << 2) - 1 + (1 << 6) - (1 << 4),
            (1 << 3) - 1 + (1 << 7) - (1 << 4) + (1 << 11) - (1 << 8)};
char c[4][1010];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 4; i++) scanf("%d", &a[i]);
  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++)
      if (c[j][i] == '.') b[i] = b[i] + (1 << j);
  }
  memset(dp, 0x3f, sizeof(dp));
  dp[0][b[0] + (b[1] << 4) + (b[2] << 8)] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < (1 << 12); j++) {
      if (dp[i][j] == inf) continue;
      if ((j & 15) == 15) {
        dp[i + 1][(j >> 4) + b[i + 3] * (1 << 8)] =
            min(dp[i][j], dp[i + 1][(j >> 4) + b[i + 3] * (1 << 8)]);
      }
      for (int k = 0; k < 3; k++) {
        for (int w = 0; w < 4 - k; w++) {
          t = j | (f[k] * (1 << w));
          dp[i][t] = min(dp[i][t], dp[i][j] + a[k]);
        }
      }
      dp[i + 1][(1 << 12) - 1] = min(dp[i + 1][(1 << 12) - 1], dp[i][j] + a[3]);
    }
  printf("%d\n", dp[n][0]);
}
