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 n, t, lim = 16;
int w[4], g[1005];
char ch[4][1005];
int num[4][4];
int s[4][4][4];
int dp[1005][16][16][16][16];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 4; i++) scanf("%d", &w[i]);
  for (int i = 0; i < 4; i++) scanf("%s", ch[i]);
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 0; i < n; i++) {
    t = 0;
    for (int j = 0; j < 4; j++) t <<= 1, t += (ch[j][i] == '.');
    g[i] = t;
  }
  for (int i = n; i < n + 5; i++) g[i] = lim - 1;
  dp[0][g[0]][g[1]][g[2]][g[3]] = 0;
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < 4 - i; j++) {
      for (int k = 0; k < 4; k++)
        for (int p = 0; p < 4; p++) num[k][p] = 0;
      for (int k = 0; k <= i; k++)
        for (int p = j; p <= j + i; p++) num[k][p] = 1;
      for (int k = 0; k < 4; k++) {
        t = 0;
        for (int p = 0; p < 4; p++) t <<= 1, t += num[k][p];
        s[i][j][k] = t;
      }
    }
  for (int i = 0; i < n; i++)
    for (int a = 0; a < lim; a++)
      for (int b = 0; b < lim; b++)
        for (int c = 0; c < lim; c++)
          for (int d = 0; d < lim; d++) {
            if (dp[i][a][b][c][d] >= 0x3f3f3f3f) continue;
            if (a == lim - 1)
              dp[i + 1][b][c][d][g[i + 4]] =
                  min(dp[i + 1][b][c][d][g[i + 4]], dp[i][a][b][c][d]);
            for (int j = 0; j < 4; j++)
              for (int k = 0; k < 4 - j; k++)
                dp[i][a | s[j][k][0]][b | s[j][k][1]][c | s[j][k][2]]
                  [d | s[j][k][3]] = min(dp[i][a | s[j][k][0]][b | s[j][k][1]]
                                           [c | s[j][k][2]][d | s[j][k][3]],
                                         dp[i][a][b][c][d] + w[j]);
          }
  printf("%d", dp[n - 1][lim - 1][lim - 1][lim - 1][lim - 1]);
  return 0;
}
