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 = 1 << 10, INF = 1e9;
int n, a[5], col_mask[N];
char s[5][N];
unordered_map<int, int> dp[N];
int solve(int pos, int mask) {
  if (dp[pos].count(mask)) return dp[pos][mask];
  if (pos == n) return 0;
  dp[pos][mask] = INF;
  if ((mask & ((1 << 4) - 1)) == 0)
    dp[pos][mask] = solve(pos + 1, mask >> 4 | col_mask[pos + 5] << 12);
  for (int sz = 1; sz <= 4; sz++) {
    for (int i = 1; i <= 4 - sz + 1; i++) {
      int new_mask = mask;
      for (int x = pos + 1; x <= pos + sz; x++)
        for (int y = i; y <= i + sz - 1; y++)
          if (mask >> (x - pos - 1) * 4 + y - 1 & 1)
            new_mask -= (1 << (x - pos - 1) * 4 + y - 1),
                dp[pos][mask] =
                    min(solve(pos, new_mask) + a[sz], dp[pos][mask]);
    }
  }
  return dp[pos][mask];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", a + i);
  for (int i = 1; i <= 4; i++) scanf("%s", s[i] + 1);
  int start_mask = 0;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= 4; j++) {
      int pos = (j - 1) * 4 + i - 1;
      start_mask += (1 << pos) * (s[i][j] == '*');
    }
  for (int col = 1; col <= n; col++) {
    for (int i = 1; i <= 4; i++)
      col_mask[col] += (1 << i - 1) * (s[i][col] == '*');
  }
  printf("%d", solve(0, start_mask));
}
