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;
const int maxN = 1010;
int dp[maxN][(1 << 12)];
const int inf = (int)1e7 + 10;
int n;
int a[5];
char c[5][maxN];
int masks[maxN];
int prec[10][10];
int rowmask[maxN];
void precalc() {
  for (int j = 1; j <= 4; j++) {
    for (int k = 1; k <= 5 - j; k++) {
      if (k == 4) continue;
      for (int x1 = j; x1 <= j + k - 1; x1++) {
        for (int x2 = 0; x2 <= k - 1; x2++) {
          prec[j][k] |= 1 << (4 * x2 + x1 - 1);
        }
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  precalc();
  for (int i = 1; i <= 4; i++) cin >> a[i];
  for (int i = 1; i < maxN; i++) {
    for (int j = 0; j < (1 << 12); j++) {
      dp[i][j] = inf;
    }
  }
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> c[i][j];
    }
  }
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      if (c[i][j] == '*') {
        rowmask[j] |= (1 << (i - 1));
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 4; j++) {
      for (int k = 0; k <= 2; k++) {
        if (c[j][i + k] == '*') {
          masks[i] |= 1 << (4 * k + (j - 1));
        }
      }
    }
  }
  dp[1][masks[1]] = 0;
  for (int i = 1; i <= n; i++) {
    for (int mask = (1 << 12) - 1; mask >= 0; mask--) {
      if ((mask & 15) == 0) {
        int to = (mask >> 4) + (rowmask[i + 3] << 8);
        dp[i + 1][to] = min(dp[i + 1][to], dp[i][mask]);
        continue;
      }
      for (int j = 1; j <= 4; j++) {
        for (int k = 1; k <= 5 - j; k++) {
          if (k == 4) {
            dp[i + 4][masks[i + 4]] =
                min(dp[i + 4][masks[i + 4]], dp[i][mask] + a[4]);
            continue;
          }
          int to = (mask | prec[j][k]) ^ prec[j][k];
          dp[i][to] = min(dp[i][to], dp[i][mask] + a[k]);
        }
      }
    }
  }
  int t1 = min(dp[n - 2][0], dp[n - 1][0]);
  int t2 = min(dp[n][0], dp[n + 1][0]);
  int t3 = min(dp[n + 2][0], dp[n + 3][0]);
  cout << min(min(t1, t2), min(t3, dp[n + 4][0]));
  return 0;
}
