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<pair<int, int> > vs;
int getCode(int x, int y) { return (1 << (3 * x + (2 - y))); }
int n, cost[5], step, col[1010], dp[4096][1010];
string str[4];
int main() {
  ios_base::sync_with_stdio(false);
  for (int i = 0; i < 4; i++) step |= getCode(i, 0);
  cin >> n;
  for (int i = 1; i <= 4; i++) cin >> cost[i];
  for (int i = 0; i < 4; i++) cin >> str[i];
  for (int j = 0; j < n; j++)
    for (int i = 0; i < 4; i++)
      if (str[i][j] == '*') col[j] |= getCode(i, 2);
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < 3; j++)
      for (int k = 1; i + k <= 4 && j + k <= 3; k++) {
        vs.push_back((make_pair((0), (cost[k]))));
        for (int x = i; x < i + k; x++)
          for (int y = j; y < j + k; y++) vs.back().first |= getCode(x, y);
      }
  for (int i = 0; i < 4096; i++) fill(dp[i], dp[i] + 1010, 2147483647);
  dp[col[0]][0] = 0;
  int res = 2147483647;
  for (int j = 0, m, c, m_; j < n; j++) {
    for (int mask = 4095; mask >= 0; mask--) {
      if (dp[mask][j] == 2147483647) continue;
      for (auto v : vs) {
        tie(m, c) = v;
        m_ = (mask | m) ^ m;
        dp[m_][j] = min(dp[m_][j], dp[mask][j] + c);
      }
      if (j >= 2) dp[0][j + 1] = min(dp[0][j + 1], dp[mask][j] + cost[4]);
      if ((mask & step) == 0)
        dp[(mask << 1) | col[j + 1]][j + 1] =
            min(dp[(mask << 1) | col[j + 1]][j + 1], dp[mask][j]);
    }
  }
  cout << min(res, dp[0][n]) << endl;
  return 0;
}
