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 MOD = 1e9 + 7;
const int MAX = 1e3 + 2;
string board[MAX];
int w[4], n;
int dp[4][MAX][4 * 4 * 4 * 4 + 2];
vector<int> tmp;
int go(int r, int c, vector<int> rem) {
  if (r == 0)
    for (int &v : rem) v = max(v - 1, 0);
  if (c == n) return 0;
  int x = 0;
  for (int v : rem) x = x * 4 + v;
  int &ans = dp[r][c][x];
  if (ans == -1) {
    ans = INT_MAX;
    int nr = r == 3 ? 0 : r + 1;
    int nc = r == 3 ? c + 1 : c;
    int lim = min(4, min(4 - r, n - c));
    for (int i = 0; i < lim; i++) {
      if (board[r][c] == '.' || rem[r] > 0) {
        ans = min(ans, go(nr, nc, rem));
      }
      vector<int> tmp = rem;
      for (int j = r; j < r + i + 1; j++) {
        tmp[j] = max(tmp[j], i + 1);
      }
      ans = min(ans, w[i] + go(nr, nc, tmp));
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(dp, -1, sizeof dp);
  cin >> n;
  for (int i = 0; i < 4; i++) cin >> w[i];
  for (int i = 0; i < n; i++) {
    cin >> board[i];
  }
  cout << go(0, 0, {0, 0, 0, 0}) << '\n';
}
