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 c, cost[6], stamp[6];
int dp[1011][(1 << 12) + 11];
char grid[6][1011];
int main() {
  cin >> c;
  for (int i = 1; i <= 4; i++) cin >> cost[i];
  stamp[1] = 1;
  stamp[2] = 51;
  stamp[3] = 1911;
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < c; j++) cin >> grid[i][j];
  memset(dp, 0x7f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < c; i++) {
    int colmask = 0;
    for (int j = 0; j < 4; j++) colmask |= ((int)(grid[j][i] == '*') << j);
    for (int mask = 0; mask < (1 << 12); mask++) {
      if (dp[i][mask] > ((1 << 29) + 123))
        continue;
      else if ((colmask & mask) == colmask)
        dp[i + 1][mask >> 4] = min(dp[i + 1][mask >> 4], dp[i][mask]);
      for (int j = 1; j <= 3; j++)
        for (int k = 0; k <= 4 - j; k++)
          dp[i][mask | (stamp[j] << k)] =
              min(dp[i][mask | (stamp[j] << k)], dp[i][mask] + cost[j]);
      dp[i + 1][(1 << 12) - 1] =
          min(dp[i + 1][(1 << 12) - 1], dp[i][mask] + cost[4]);
    }
  }
  int ans = ((1 << 29) + 123);
  for (int i = 0; i < (1 << 12); i++) ans = min(ans, dp[c][i]);
  printf("%d\n", ans);
  return 0;
}
