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 Maxd = 5;
const int Maxn = 1010;
const int Maxm = 1 << 16;
const int Inf = 1000000000;
int n;
int cost[Maxd];
string B[Maxd];
int rmask[Maxn];
vector<int> masks[Maxd];
int cur, dp[2][Maxm];
int getId(int r, int c) { return r * 4 + c; }
int addRow(int mask, int row) {
  mask >>= 4;
  if (row < n) mask |= rmask[row] << 12;
  return mask;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < Maxd; i++) {
    scanf("%d", &cost[i]);
    for (int j = 0; j + i <= 4; j++) {
      int mask = (Maxm - 1);
      for (int r = 0; r < i; r++)
        for (int c = j; c < j + i; c++) mask ^= 1 << getId(r, c);
      masks[i].push_back(mask);
    }
  }
  for (int i = 0; i < 4; i++) cin >> B[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 4; j++) rmask[i] |= int(B[j][i] == '*') << j;
  fill((int*)dp, (int*)dp + 2 * Maxm, Inf);
  int my = 0;
  for (int i = 0; i < 4; i++) my = addRow(my, i);
  dp[cur][my] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = Maxm - 1; j >= 0; j--)
      if (dp[cur][j] < Inf) {
        if (!(j & 15)) {
          int nj = addRow(j, i + 4);
          dp[!cur][nj] = min(dp[!cur][nj], dp[cur][j]);
        }
        for (int k = 1; k < Maxd && i + k <= n; k++)
          for (int z = 0; z < masks[k].size(); z++) {
            int nj = (j & masks[k][z]);
            dp[cur][nj] = min(dp[cur][nj], dp[cur][j] + cost[k]);
          }
        dp[cur][j] = Inf;
      }
    cur = !cur;
  }
  printf("%d\n", dp[cur][0]);
  return 0;
}
