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 << 16];
char art[7][1008];
int n;
int c[5];
vector<int> modi[1008];
void upd(int &x, int y) { x = min(x, y); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= 4; i++) scanf("%s", art[i] + 1);
  int now = 0;
  memset(dp[now], 0x3f, sizeof dp[now]);
  dp[now][0] = 0;
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= 4 - i + 1; j++) {
      int mas = (1 << 16) - 1;
      for (int x = 0; x < i; x++)
        for (int y = j - 1; y < j - 1 + i; y++) {
          mas ^= (1 << (x * 4 + y));
        }
      modi[i].push_back(mas);
    }
  }
  for (int i = 1; i <= n + 4; i++) {
    now ^= 1;
    memset(dp[now], 0x3f, sizeof dp[now]);
    int nm = 0;
    for (int j = 1; j <= 4; j++)
      if (art[j][i] == '*') nm |= (1 << 11 + j);
    for (int j = (1 << 16) - 1; j >= 0; j--)
      if (dp[now ^ 1][j] < 1e9) {
        for (int k = 1; k <= 4; k++)
          for (int l = 0; l < modi[k].size(); l++)
            upd(dp[now ^ 1][j & modi[k][l]], dp[now ^ 1][j] + c[k]);
        if (j & 15) continue;
        upd(dp[now][(j >> 4) | nm], dp[now ^ 1][j]);
      }
  }
  printf("%d\n", dp[now][0]);
  return 0;
}
