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;
int n;
int a[5];
char g[4][1001];
int dp[4][1001][16][16][16];
int solve(int i, int j, int m1, int m2, int m3) {
  if (i == 4) return solve(0, j + 1, m2, m3, 0);
  if (j == n) return 0;
  int &ret = dp[i][j][m1][m2][m3];
  if (ret != -1) return ret;
  ret = 1 << 30;
  if (g[i][j] == '.' || ((1 << i) & m1) != 0)
    ret = min(ret, solve(i + 1, j, m1, m2, m3));
  if (j + 4 <= n && i == 0) ret = min(ret, a[4] + solve(0, j + 1, 15, 15, 15));
  for (int k = 1; k <= 3; k++) {
    if (n - j < k || 4 - i < k) break;
    int M1 = m1, M2 = m2, M3 = m3;
    for (int c = 0; c < k; c++) {
      int cc = i + c;
      M1 |= (1 << cc);
      if (k > 1) M2 |= (1 << cc);
      if (k > 2) M3 |= (1 << cc);
    }
    ret = min(ret, a[k] + solve(i + 1, j, M1, M2, M3));
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 1; i <= 4; i++) cin >> a[i];
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < n; j++) cin >> g[i][j];
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0, 0, 0, 0);
  return 0;
}
