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 oo = 1e9;
int n, cost[5], col[1001], dp[5][1002][1 << 12 + 1], fl[5][5];
string g[4];
int calc(int r, int c, int msk) {
  if (c == n) return 0;
  if (r == 4) {
    if ((msk & col[c]) != col[c]) return 1e9;
    return calc(0, c + 1, msk >> 4);
  }
  int &ret = dp[r][c][msk];
  if (ret != -1) return ret;
  ret = calc(r + 1, c, msk);
  for (int s = 1; s <= 4 && c + s <= n && r + s <= 4; ++s) {
    if (s == 4)
      ret = min(ret, cost[s] + calc(0, c + 1, (1 << 12) - 1));
    else {
      ret = min(ret, cost[s] + calc(r + 1, c, msk | fl[r][s]));
    }
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%d", &n);
  for (int i = 1; i <= 4; ++i) scanf("%d", &cost[i]);
  for (int i = 0; i < 4; ++i) {
    cin >> g[i];
  }
  for (int r = 0; r < 4; ++r) {
    for (int s = 1; s <= 3 && s + r <= 4; ++s) {
      for (int d1 = 0; d1 < s; ++d1) {
        for (int d2 = 0; d2 < s; ++d2) {
          int nwR = d1 + r;
          fl[r][s] |= (1 << (d2 * 4 + nwR));
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 4; ++j) {
      if (g[j][i] == '*') col[i] = col[i] | (1 << j);
    }
  }
  printf("%d\n", calc(0, 0, 0));
  return 0;
}
