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;
char s[1005];
int n, ans, cnt, msk, f[1005][(1 << 16) + 2], a[6][1005], c[10], g[10];
int rev(int x) { return x ^ ((1 << 16) - 1); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= 4; i++) {
    scanf("%s", s + 1);
    for (int j = 1; j <= n; j++) a[i][j] = (s[j] == '*');
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < (1 << 16); j++) f[i][j] = 1e9 + 7;
  cnt = -1;
  msk = 0;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= 4; j++) {
      cnt++;
      if (a[j][i]) msk |= (1 << cnt);
    }
  f[0][msk] = 0;
  g[0] = 0;
  g[1] = 1;
  g[2] = (1 << 0) ^ (1 << 1) ^ (1 << 4) ^ (1 << 5);
  g[3] = (1 << 0) ^ (1 << 1) ^ (1 << 2) ^ (1 << 4) ^ (1 << 5) ^ (1 << 6) ^
         (1 << 8) ^ (1 << 9) ^ (1 << 10);
  g[4] = (1 << 16) - 1;
  for (int k = 0; k <= n; k++) {
    int del = 0;
    for (int j = 1; j <= 4; j++)
      if (a[j][k + 5]) del |= (1 << (j + 11));
    for (int j = 0; j < (1 << 16); j++) {
      if (f[k][j] >= 1e9 + 7) continue;
      for (int aa = 0; aa <= 4; aa++)
        for (int bb = 0; bb <= 3; bb++)
          for (int cc = 0; cc <= 2; cc++)
            for (int dd = 0; dd <= 1; dd++) {
              int cnt = rev(g[aa]) & rev(g[bb] << 1) & rev(g[cc] << 2) &
                        rev(g[dd] << 3);
              if ((cnt & j & 15) == 0) {
                int nw = cnt & j;
                nw >>= 4;
                nw ^= del;
                f[k + 1][nw] =
                    min(f[k + 1][nw], f[k][j] + c[aa] + c[bb] + c[cc] + c[dd]);
              }
            }
    }
  }
  ans = 1e9 + 7;
  for (int i = n - 4; i <= n; i++) ans = min(ans, f[i][0]);
  printf("%d\n", ans);
  return 0;
}
