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;
void testCase() {}
string grid[4];
int dp[1111][5][5][5][5];
bool done[1111][5][5][5][5];
int a1, a2, a3, a4;
int solve(int ind, vector<int> v) {
  if (ind < 0) return 0;
  if (done[ind][v[0]][v[1]][v[2]][v[3]]) return dp[ind][v[0]][v[1]][v[2]][v[3]];
  done[ind][v[0]][v[1]][v[2]][v[3]] = true;
  for (int i = 0; i < 4; i++) {
    if (v[i] == 0) {
      if (grid[i][ind] == '.') {
        vector<int> w = v;
        w[i]++;
        dp[ind][v[0]][v[1]][v[2]][v[3]] = solve(ind, w);
        return dp[ind][v[0]][v[1]][v[2]][v[3]];
      }
      v[i]++;
      int best = a1 + solve(ind, v);
      v[i]--;
      vector<int> w;
      w.resize(4);
      w[0] = max(2, v[0]);
      w[1] = max(2, v[1]);
      w[2] = v[2];
      w[3] = v[3];
      if (w != v) best = min(best, a2 + solve(ind, w));
      w.clear();
      w.resize(4);
      w[0] = v[0];
      w[1] = max(2, v[1]);
      w[2] = max(2, v[2]);
      w[3] = v[3];
      if (w != v) best = min(best, a2 + solve(ind, w));
      w.clear();
      w.resize(4);
      w[0] = v[0];
      w[1] = v[1];
      w[2] = max(2, v[2]);
      w[3] = max(2, v[3]);
      if (w != v) best = min(best, a2 + solve(ind, w));
      w.clear();
      w.resize(4);
      w[0] = v[0];
      w[1] = max(3, v[1]);
      w[2] = max(3, v[2]);
      w[3] = max(3, v[3]);
      if (w != v) best = min(best, a3 + solve(ind, w));
      w.clear();
      w.resize(4);
      w[0] = max(3, v[0]);
      w[1] = max(3, v[1]);
      w[2] = max(3, v[2]);
      w[3] = v[3];
      if (w != v) best = min(best, a3 + solve(ind, w));
      w[0] = 4;
      w[1] = 4;
      w[2] = 4;
      w[3] = 4;
      if (w != v) best = min(best, a4 + solve(ind, w));
      dp[ind][v[0]][v[1]][v[2]][v[3]] = best;
      return best;
    }
  }
  for (int i = 0; i < 4; i++) v[i]--;
  dp[ind][v[0] + 1][v[1] + 1][v[2] + 1][v[3] + 1] = solve(ind - 1, v);
  ;
  return dp[ind][v[0] + 1][v[1] + 1][v[2] + 1][v[3] + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  cin >> a1 >> a2 >> a3 >> a4;
  for (int i = 0; i < 4; i++) cin >> grid[i];
  vector<int> v;
  v = {0, 0, 0, 0};
  cout << solve(n - 1, v) << "\n";
}
