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 = 1005;
int n, A[4], pos[4], f[N][4][4][4][4];
char s[4][N];
void dfs(int L, int p, int a[4], int v) {
  if (p == 4) {
    for (int i = 0; i < 4; i++) a[i] = max(a[i], 0);
    f[L][a[0]][a[1]][a[2]][a[3]] = min(f[L][a[0]][a[1]][a[2]][a[3]], v);
    return;
  }
  if (s[p][L] == '.' || a[p] >= 0) dfs(L, p + 1, a, v);
  int b[4];
  for (int i = 0; p + i < 4; i++) {
    for (int j = 0; j < 4; j++) b[j] = a[j];
    for (int j = p; j <= i + p; j++) b[j] = max(b[j], i);
    dfs(L, p + 1, b, v + A[i]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 4; i++) scanf("%d", &A[i]);
  for (int i = 0; i < 4; i++) scanf("%s", s[i] + 1);
  memset(f, 0x3f, sizeof(f));
  f[0][0][0][0][0] = 0;
  for (int i = 0; i < n; i++)
    for (int a = 0; a < 4; a++)
      for (int b = 0; b < 4; b++)
        for (int c = 0; c < 4; c++)
          for (int d = 0; d < 4; d++)
            if (f[i][a][b][c][d] < 1e9) {
              pos[0] = a - 1;
              pos[1] = b - 1;
              pos[2] = c - 1;
              pos[3] = d - 1;
              dfs(i + 1, 0, pos, f[i][a][b][c][d]);
            }
  printf("%d\n", f[n][0][0][0][0]);
}
