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, tmp = (1 << 16) - 1, inf = 0x3f3f3f3f;
int n, a[6], l[N], bin[20], num[6][6], dp[N][tmp + 100];
vector<int> mark[6];
char c[6][N];
int ch(int size, int r) {
  int ans = 0;
  for (int i = 1; i <= size; i++)
    for (int j = 1; j <= size; j++) ans |= bin[num[r + i - 1][j]];
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= 4; i++) scanf("%s", c[i] + 1);
  bin[1] = 1;
  for (int i = 2; i <= 16; i++) bin[i] = bin[i - 1] << 1;
  int sz = 0, be = 0;
  for (int j = 1; j <= 4; j++)
    for (int i = 1; i <= 4; i++) num[i][j] = ++sz;
  for (int j = 1; j <= n; j++)
    for (int i = 1; i <= 4; i++)
      if (c[i][j] == '*') l[j] |= bin[i];
  for (int k = 1; k <= 4; k++) {
    for (int i = 1; i + k - 1 <= 4; i++) mark[k].push_back(tmp ^ ch(k, i));
  }
  memset(dp, inf, sizeof dp);
  for (int i = 1; i <= min(4, n); i++) be |= (l[i] << (i - 1) * 4);
  dp[1][be] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = tmp; j >= 0; j--) {
      if (dp[i][j] == inf) continue;
      if (!(j & 15)) {
        int t = (j >> 4) | (l[i + 4] << 12);
        dp[i + 1][t] = min(dp[i + 1][t], dp[i][j]);
      }
      for (int k = 1; i + k - 1 <= n; k++)
        for (int q = 0; q < mark[k].size(); q++) {
          int s = mark[k][q];
          if ((j & s) != j) {
            dp[i][j & s] = min(dp[i][j & s], dp[i][j] + a[k]);
          }
        }
    }
  cout << dp[n][0] << endl;
}
