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 = 1005, M = (1 << 16) - 1;
int n, S, inf;
int a[10], b[N], f[N][M + 7], Mk[4][4];
char s[4][N];
void make(int l, int x) {
  Mk[x][l] = M;
  for (int i = 0; i < x; i++)
    for (int j = 0; j < x; j++) Mk[x][l] ^= 1 << (j * 4 + l + i);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
  for (int i = 0; i < 4; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 4; j++)
      if (s[j][i] == '*') b[i] |= 1 << j;
  memset(f, 60, sizeof f);
  inf = f[0][0];
  for (int i = 1; i <= 4; i++) {
    for (int j = 0; i + j <= 4; j++) make(j, i);
  }
  for (int i = 0; i < 4; i++) S |= b[i + 1] << (i * 4);
  f[1][S] = 0;
  for (int i = 1; i <= n; i++) {
    for (int S = M; ~S; S--)
      if (f[i][S] != inf) {
        if ((S & 15) == 0) {
          int SS = (S >> 4) | (b[i + 4] << 12);
          f[i + 1][SS] = min(f[i + 1][SS], f[i][S]);
        }
        for (int j = 1; j <= 4 && i + j <= n + 1; j++) {
          for (int k = 0; j + k <= 4; k++) {
            int SS = S & Mk[j][k];
            f[i][SS] = min(f[i][SS], f[i][S] + a[j]);
          }
        }
      }
  }
  printf("%d\n", f[n + 1][0]);
}
