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 N = 1002;
int i, j, d, x, y, s, a[4], tmp, f[N][1 << 12], n, st[N];
char s1[4][N];
vector<int> v[4];
inline void M(int &x, int y) {
  if (y < x) x = y;
}
int main() {
  scanf("%d", &n);
  for (i = 0; i < 4; i++) scanf("%d", &a[i]);
  for (i = 0; i < 4; i++) scanf("%s", s1[i]);
  for (i = 0; i < n; i++)
    for (j = 0; j < 4; j++) st[i] = st[i] << 1 | (s1[j][i] == '*');
  for (i = 0; i < 3; i++)
    for (j = 0; j < 4; j++)
      for (d = 0; i + d < 3 && j + d < 4; d++) {
        tmp = (1 << 12) - 1;
        for (x = i; x <= i + d; x++)
          for (y = j; y <= j + d; y++) tmp ^= 1 << x * 4 + y;
        v[d].push_back(tmp);
      }
  memset(f, 63, sizeof(f));
  f[2][st[2] | st[1] << 4 | st[0] << 8] = 0;
  for (i = 2; i < n; i++)
    for (s = (1 << 12) - 1; ~s; s--)
      if (f[i][s] < 1e9) {
        if (s < (1 << 8)) M(f[i + 1][s << 4 | st[i + 1]], f[i][s]);
        M(f[i + 1][0], f[i][s] + a[3]);
        for (d = 0; d < 3; d++)
          for (j = 0; j < v[d].size(); j++)
            M(f[i][s & v[d][j]], f[i][s] + a[d]);
      }
  printf("%d", f[n - 1][0]);
}
