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 = 999999999;
const double PI = 3.1415931;
const double eps = 1e-9;
const int maxN = 1000 + 10;
char mat[6][maxN];
int dp[maxN][16][16][16], n;
int c1, c2, c3, c4, b[10][maxN], a[maxN];
int mem(int pos, int a1, int a2, int a3) {
  if (pos <= 1 && a1 == 0) {
    return 0;
  }
  if (a1 == 0) return mem(pos - 1, a2, a3, ((pos - 3 > 0) ? a[pos - 3] : 0));
  int &ret = dp[pos][a1][a2][a3];
  if (ret != -1) return ret;
  ret = oo;
  ret = min(ret, c1 + mem(pos, a1 & 14, a2, a3));
  ret = min(ret, c1 + mem(pos, a1 & 13, a2, a3));
  ret = min(ret, c1 + mem(pos, a1 & 11, a2, a3));
  ret = min(ret, c1 + mem(pos, a1 & 7, a2, a3));
  ret = min(ret, c2 + mem(pos, a1 & 3, a2 & 3, a3));
  ret = min(ret, c2 + mem(pos, a1 & 9, a2 & 9, a3));
  ret = min(ret, c2 + mem(pos, a1 & 12, a2 & 12, a3));
  ret = min(ret, c3 + mem(pos, a1 & 1, a2 & 1, a3 & 1));
  ret = min(ret, c3 + mem(pos, a1 & 8, a2 & 8, a3 & 8));
  ret = min(ret, c4 + mem(pos - 4, ((pos - 4 > 0) ? a[pos - 4] : 0),
                          ((pos - 5 > 0) ? a[pos - 5] : 0),
                          ((pos - 6 > 0) ? a[pos - 6] : 0)));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  memset(dp, -1, sizeof dp);
  cin >> n;
  cin >> c1 >> c2 >> c3 >> c4;
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> mat[i][j];
      if (mat[i][j] == '*') b[i][j] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    a[i] = 8 * b[4][i] + 4 * b[3][i] + 2 * b[2][i] + b[1][i];
  }
  cout << mem(n, a[n], a[n - 1], a[n - 2]);
  return 0;
}
