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 d[1100][1 << 12], m[1100][4], a[5];
vector<int> s[3] = {{1, 2, 4, 8}, {51, 102, 204}, {-2185, -274}};
int main() {
  int N;
  cin >> N;
  for (int i = 0; i < 4; i++) cin >> a[i];
  for (int r = 0; r < 4; r++)
    for (int c = 0; c < N; c++) {
      char x;
      cin >> x;
      m[c][r] = x == '*';
    }
  for (int i = N + 5; i >= 0; i--) {
    for (int b = 0; b < 1 << 12; b++) {
      if (b % (1 << 4) == 0) {
        int c = b / 16;
        for (int r = 0; r < 4; r++) c += (1 << (8 + r)) * m[i][r];
        d[i][b] = d[i + 1][c];
        continue;
      }
      d[i][b] = d[i + 1][0] + a[3];
      for (int l = 0; l < 3; l++)
        for (int S : s[l]) d[i][b] = min(d[i][b], a[l] + d[i][b & (~S)]);
    }
  }
  cout << d[0][0];
}
