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 dp[1010][630], a[5], n;
string s[4];
int encode(vector<int> p) { return ((p[3] * 5 + p[2]) * 5 + p[1]) * 5 + p[0]; }
vector<int> decode(int x) { return {x % 5, x / 5 % 5, x / 25 % 5, x / 125}; }
bool Min(int &a, int b) { return a > b ? a = b, 1 : 0; }
int main() {
  cin >> n;
  for (int i = (1); i <= (4); i++) cin >> a[i];
  for (int i = (0); i <= (3); i++) cin >> s[i];
  memset(dp, 0x3f, sizeof(dp)), dp[0][0] = 0;
  for (int i = (0); i <= (n - 1); i++)
    for (int k = (0); k <= (624); k++) {
      if (dp[i][k] == 0x3f3f3f3f) continue;
      vector<int> cur = decode(k), nxt;
      nxt = cur;
      bool ok = true;
      for (int j = (0); j <= (3); j++) {
        if (nxt[j] == 0 && s[j][i] == '*') ok = false;
        nxt[j] = max(nxt[j] - 1, 0);
      }
      if (ok) Min(dp[i + 1][encode(nxt)], dp[i][k]);
      for (int len = (1); len <= (4); len++)
        if (i + len - 1 <= n)
          for (int j = (0); j <= (4 - len); j++) {
            nxt = cur;
            for (int l = (0); l <= (len - 1); l++)
              nxt[j + l] = max(nxt[j + l], len);
            Min(dp[i][encode(nxt)], dp[i][k] + a[len]);
          }
    }
  cout << dp[n][0] << endl;
  return 0;
}
