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;
vector<int> mark[4];
string sins[4];
int cost[5];
int sta[1010];
int dp[1010][0xffff + 1];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < 4; i++) cin >> cost[i];
  for (int i = 0; i < 4; i++) cin >> sins[i];
  for (int j = 0; j < n; j++) {
    sta[j] = 0;
    for (int i = 0; i < 4; i++) {
      sta[j] |= (sins[i][j] == '*') ? (1 << i) : 0;
    }
  }
  int S = 0;
  for (int i = 0; i < 4; i++) {
    S |= sta[i] << (i * 4);
    sta[n + i] = 0;
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 3; j >= i; j--) {
      int t = 0xffff;
      for (int k = 0; k <= i; k++) {
        for (int y = j; y >= j - i; y--) t ^= 1 << (4 * k + y);
      }
      mark[i].push_back(t);
    }
  }
  memset(dp, 0x7f, sizeof(dp));
  dp[0][S] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0xffff; j >= 0; j--) {
      if (dp[i][j] == 0x7f7f7f7f) continue;
      if ((j & 0xf) == 0) {
        int S = (((j & 0xffff) >> 4) | (0xf000 & (sta[i + 4] << 12)));
        dp[i + 1][S] = min(dp[i + 1][S], dp[i][j]);
      }
      for (int k = 0; k < 4 && k + i < n; k++) {
        vector<int>::iterator y;
        for (y = mark[k].begin(); y != mark[k].end(); y++) {
          dp[i][j & (*y)] = min(dp[i][j & (*y)], dp[i][j] + cost[k]);
        }
      }
    }
  }
  cout << dp[n][0] << endl;
}
