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 Y[5];
int dp[66000][1001];
int B[14] = {0, 3, 6, 12, 15, 119, 238, 4095, 7, 14, 255, 15, 127, 239};
int D[14] = {0};
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i < 5; i++) cin >> Y[i];
  char A[4][n];
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) cin >> A[i][j];
  }
  for (int i = 1; i < 4; i++) {
    D[i] = Y[2];
  }
  D[4] = 2 * Y[2];
  D[5] = Y[3];
  D[6] = Y[3];
  D[7] = Y[4];
  D[8] = 2 * Y[2];
  D[9] = 2 * Y[2];
  D[10] = 2 * Y[3];
  D[11] = 3 * Y[2];
  D[12] = Y[2] + Y[3];
  D[13] = D[12];
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 1 << 12; j++) {
      dp[j][i] = 1e9;
    }
  }
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 1 << 12; j++) {
      for (int k = 0; k < 14; k++) {
        int lol = j;
        lol |= B[k];
        int a = (j / 16);
        a |= B[k];
        int val = 0;
        for (int l = 0; l < 4; l++) {
          if (A[l][i] == '*') {
            int uu = (1 << l) & lol;
            if (uu == 0) {
              val += Y[1];
            }
          }
        }
        val += D[k];
        dp[a][i + 1] = min(dp[a][i + 1], dp[j][i] + val);
      }
    }
  }
  int ans = 1e9;
  for (int i = 0; i < 1 << 12; i++) {
    ans = min(dp[i][n], ans);
  }
  cout << ans;
}
int main() {
  int t;
  t = 1;
  while (t--) solve();
}
