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 N = 1010, INF = 1e9 + 10;
char ch[N];
int v[6], c[6][6], sta[N], f[2][1 << 17];
int main() {
  for (register int i = 1; i <= 4; i++) {
    for (register int j = 1; j <= 4 - i + 1; j++) {
      for (register int t = 1; t <= j; t++) {
        c[i][j] <<= 4;
        for (int k = i; k <= i + j - 1; k++) c[i][j] += (1 << (k - 1));
      }
      for (register int t = j + 1; t <= 4; t++) c[i][j] <<= 4;
    }
  }
  int n, pre = 0;
  scanf("%d", &n);
  for (register int i = 1; i <= 4; i++) scanf("%d", &v[i]);
  for (register int i = 1; i <= 4; i++) {
    scanf("%s", ch);
    for (register int j = 0; j < n; j++)
      sta[j] = (sta[j] << 1) + (ch[j] == '.');
  }
  for (register int i = n; i <= n + 4; i++) sta[i] = (1 << 4) - 1;
  for (register int i = 0; i <= 3; i++) pre = (pre << 4) | sta[i];
  for (register int s = 0; s < (1 << 16); s++) f[0][s] = INF;
  f[0][pre] = 0;
  for (register int i = 0; i < n; i++) {
    int now = i & 1, cur = (i + 1) & 1;
    for (register int s = 0; s < (1 << 16); s++) f[cur][s] = INF;
    for (register int s = 0; s < (1 << 16); s++) {
      if (f[now][s] == INF) continue;
      if ((s >> 12) == 15) {
        int s2 = ((s & ((1 << 12) - 1)) << 4) | sta[i + 4];
        f[cur][s2] = min(f[cur][s2], f[now][s]);
      }
      for (register int k = 1; k <= 4; k++)
        for (register int j = 1; j <= 4 - k + 1; j++)
          f[now][s | c[k][j]] = min(f[now][s | c[k][j]], f[now][s] + v[j]);
    }
  }
  printf("%d\n", f[(n - 1) & 1][(1 << 16) - 1]);
  return 0;
}
