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 mask[5][5], n, a[5], dp[1010][1 << 16], b[5][2005];
char mp[5][2005];
void premask() {
  mask[0][0] = 0b1111111111111111;
  mask[0][1] = 0b0111111111111111;
  mask[1][0] = 0b1111111111111111;
  mask[1][1] = 0b1011111111111111;
  mask[1][2] = 0b0011001111111111;
  mask[2][0] = 0b1111111111111111;
  mask[2][1] = 0b1101111111111111;
  mask[2][2] = 0b1001100111111111;
  mask[2][3] = 0b0001000100011111;
  mask[3][0] = 0b1111111111111111;
  mask[3][1] = 0b1110111111111111;
  mask[3][2] = 0b1100110011111111;
  mask[3][3] = 0b1000100010001111;
  mask[3][4] = 0b0000000000000000;
}
int main() {
  cin >> n;
  for (int i = 1; i <= 4; i++) cin >> a[i];
  for (int i = 0; i < 4; i++) scanf("%s", mp[i]);
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < n; j++) b[i][j] = mp[i][j] == '*' ? 1 : 0;
  memset(dp, 0x3f, sizeof dp);
  int init = 0;
  for (int j = 0; j < 4; j++)
    for (int i = 0; i < 4; i++) init <<= 1, init |= b[i][j];
  dp[0][init] = 0;
  premask();
  for (int i = 0; i < n; i++) {
    for (int s = 0; s < (1 << 16); s++) {
      if (dp[i][s] == 0x3f3f3f3f) continue;
      for (int l1 = 0; l1 <= 1; l1++)
        for (int l2 = 0; l2 <= 2; l2++)
          for (int l3 = 0; l3 <= 3; l3++)
            for (int l4 = 0; l4 <= 4; l4++) {
              int now = s;
              now &= mask[0][l1];
              now &= mask[1][l2];
              now &= mask[2][l3];
              now &= mask[3][l4];
              if ((now >> 15) & 1) continue;
              if ((now >> 14) & 1) continue;
              if ((now >> 13) & 1) continue;
              if ((now >> 12) & 1) continue;
              int cost = a[l1] + a[l2] + a[l3] + a[l4];
              int nxt = now;
              for (int j = 0; j < 4; j++) {
                nxt <<= 1;
                nxt &= (1 << 16) - 1;
                nxt |= b[j][i + 4];
              }
              dp[i + 1][nxt] = min(dp[i + 1][nxt], dp[i][s] + cost);
            }
    }
  }
  cout << dp[n][0] << " ";
}
