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 advanceMask(int i, vector<string>& a, int nm, int n) {
  nm >>= 4;
  for (int k = 0; k < (int)(4); ++k) {
    if (i + 2 >= n || a[k][i + 2] == '.') nm += (1 << (8 + k));
  }
  return nm;
}
int main() {
  int n;
  cin >> n;
  vector<long long> c(4);
  for (int i = 0; i < (int)(4); ++i) cin >> c[i];
  vector<string> a(4);
  for (int i = 0; i < (int)(4); ++i) cin >> a[i];
  vector<vector<vector<long long>>> dp(
      n + 1,
      vector<vector<long long>>(4, vector<long long>(1 << 12, 1000000000)));
  dp[n][0][(1 << 12) - 1] = 0;
  for (int i = (n)-1; i >= 0; --i)
    for (int j = (4) - 1; j >= 0; --j)
      for (int m = 0; m < (int)((1 << 12)); ++m) {
        int ni = i, nj = j + 1, nm = m;
        if (nj == 4) {
          nj = 0;
          ni++;
          nm = advanceMask(ni, a, nm, n);
        }
        if (m & (1 << j)) {
          dp[i][j][m] = min(dp[i][j][m], dp[ni][nj][nm]);
        }
        for (int d = 1; d <= min(4 - j, n - i); d++) {
          if (d == 4) {
            dp[i][j][m] =
                min(dp[i][j][m], dp[i + 1][0][(1 << 12) - 1] + c[d - 1]);
          } else {
            int pm = m;
            for (int x = j; x < j + d; x++) {
              for (int y = 0; y < d; y++) {
                pm |= (1 << (x + 4 * y));
              }
            }
            if (ni > i) pm = advanceMask(ni, a, pm, n);
            dp[i][j][m] = min(dp[i][j][m], dp[ni][nj][pm] + c[d - 1]);
          }
        }
      }
  int sm = 0;
  for (int i = 0; i < (int)(4); ++i)
    for (int j = 0; j < (int)(3); ++j) {
      if (j > n || a[i][j] == '.') sm += (1 << (i + j * 4));
    }
  cout << dp[0][0][sm] << endl;
}
