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;
int n, a[4], memo[1004][4][1 << 4][1 << 4][1 << 4];
string s[1004];
int dp(int x, int y, int m1, int m2, int m3) {
  if (x == n) return 0;
  int &ret = memo[x][y][m1][m2][m3];
  if (ret != -1) return ret;
  bool yes = (s[x][y] == '.');
  yes |= (m1 & (1 << (3 - y)));
  if (y == 3) return ret = dp(x + 1, 0, m2, m3, 0) + (!yes) * a[0];
  ret = 1e9;
  if (yes) ret = dp(x, y + 1, m1, m2, m3);
  for (int i = 0; i < 4 - y; i++) {
    if (x + i >= n) break;
    if (i == 3) ret = min(ret, dp(x + 4, 0, 0, 0, 0) + a[3]);
    int b = (1 << (i + 1)) - 1;
    b <<= ((3 - y) - i);
    int q1 = (m1 | b);
    int q2 = m2;
    int q3 = m3;
    if (i > 0) q2 |= b;
    if (i > 1) q3 |= b;
    ret = min(ret, dp(x, y + 1, q1, q2, q3) + a[i]);
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < 4; i++) cin >> a[i];
  string s1[4];
  for (int i = 0; i < 4; i++) cin >> s1[i];
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) s[j] += s1[i][j];
  }
  for (int i = 0; i < n; i++) reverse(s[i].begin(), s[i].end());
  memset(memo, -1, sizeof memo);
  cout << dp(0, 0, 0, 0, 0);
  return 0;
}
