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 main() {
  int n;
  cin >> n;
  vector<int> A(4);
  for (int i = 0; i < 4; ++i) cin >> A[i];
  vector<int> B(n + 4);
  for (int i = 0; i < 4; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < n; ++j) B[j] |= (s[j] == '*') << i;
  }
  const int m = 1 << 4;
  const long long INF = 1e9;
  array<long long, m * m * m> dp;
  fill(dp.begin(), dp.end(), INF);
  dp[0] = 0;
  for (int i = 0; i < n + 4; ++i) {
    long long mi = INF;
    for (int j = m - 1; j >= 0; --j) {
      for (int k = m - 1; k >= 0; --k) {
        for (int l = m - 1; l >= 0; --l) {
          if (dp[j * m * m + k * m + l] >= INF) continue;
          for (int s = 1; s <= 3; ++s) {
            for (int t = 0; t < 4; ++t) {
              if (t + s > 4) break;
              int mask = (1 << 4) - 1;
              for (int u = 0; u < s; ++u) mask ^= (1 << (u + t));
              vector<int> T = {j, k, l};
              for (int u = 0; u < s; ++u) T[u] &= mask;
              int idx = 0;
              for (auto e : T) {
                idx *= m;
                idx += e;
              }
              dp[idx] = min(dp[idx], dp[j * m * m + k * m + l] + A[s - 1]);
              mi = min(mi, dp[j * m * m + k * m + l]);
            }
          }
        }
      }
    }
    array<long long, m * m * m> dp_;
    fill(dp_.begin(), dp_.end(), INF);
    for (int j = 0; j < m; ++j) {
      for (int k = 0; k < m; ++k) {
        dp_[j * m * m + k * m + B[i]] =
            min(dp[j * m + k], dp_[j * m * m + k * m + B[i]]);
      }
    }
    dp_[0] = min(dp_[0], mi + A[3]);
    swap(dp, dp_);
  }
  cout << dp[0] << endl;
}
