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 dp[2][1 << 15], mat[5], w[5], n;
bool s[1010][5];
char ch[1010];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &w[i]);
  for (int i = 1; i <= 4; i++) {
    scanf("%s", ch);
    for (int j = 0; j < n; j++) s[j + 1][i] = ch[j] == '.';
  }
  int s0 = 0;
  mat[1] = 1;
  mat[2] = 51;
  mat[3] = 1911;
  mat[4] = 65535;
  memset(dp, 0x3f3f3f3f, sizeof(dp));
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= 3; j++)
      if (s[j][i]) s0 |= 1 << ((j - 1) * 4 + i - 1);
  int cur = 1, pre = 0;
  dp[pre][s0] = 0;
  w[0] = 0;
  for (int i = 0; i < n; i++) {
    int nxt = 0;
    memset(dp[cur], 0x3f3f3f3f, sizeof(dp[cur]));
    for (int j = 1; j <= 4; j++)
      if (s[i + 4][j]) nxt |= 1 << j - 1;
    for (int st = 0; st < (1 << 12); st++) {
      if (dp[pre][st] == 0x3f3f3f3f) continue;
      for (int a = 0; a <= 4; a++)
        for (int b = 0; b <= 3; b++)
          for (int c = 0; c <= 2; c++)
            for (int d = 0; d <= 1; d++) {
              int cost = w[a] + w[b] + w[c] + w[d];
              int add = mat[a] << (4 - a) | mat[b] << (3 - b) |
                        mat[c] << (2 - c) | mat[d] << (1 - d);
              if (((st | add) & 15) == 15) {
                int nxt_st = (st >> 4) | (add >> 4) | (nxt << 8);
                dp[cur][nxt_st] = min(dp[cur][nxt_st], dp[pre][st] + cost);
              }
            }
    }
    swap(cur, pre);
  }
  cout << dp[pre][0] << endl;
  return 0;
}
