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 mx = 1e3 + 6;
const int mm = (1 << 16) - 1;
const int INF = 0x3f3f3f3f;
int n, a[4], st[mx];
char s[4][mx];
int d[mx][mm + 6];
vector<int> mark[4];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 4; i++) scanf("%d", &a[i]);
  for (int i = 0; i < 4; i++) scanf("%s", s[i]);
  for (int j = 0; j < n; j++) {
    for (int i = 0; i < 4; i++) {
      st[j] |= (s[i][j] == '*') << i;
    }
  }
  memset(d, 0x3f, sizeof(d));
  int S = 0;
  for (int i = 0; i < 4; i++) S |= st[i] << (i * 4);
  d[0][S] = 0;
  for (int i = 1; i <= 4; i++) {
    for (int j = 0; j <= 4 - i; j++) {
      mark[i - 1].push_back(mm);
      int& SS = mark[i - 1].back();
      for (int k = j; k < j + i; k++) {
        for (int g = 0; g < i; g++) {
          SS ^= 1 << (g * 4 + k);
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = mm; j >= 0; j--) {
      if (d[i][j] == INF) continue;
      if (!(j & 15)) {
        int SS = j >> 4 | (st[i + 4] << 12);
        d[i + 1][SS] = min(d[i][j], d[i + 1][SS]);
      }
      for (int k = 1; k <= 4 && k + i <= n; k++) {
        for (auto p : mark[k - 1]) {
          d[i][j & p] = min(d[i][j & p], d[i][j] + a[k - 1]);
        }
      }
    }
  }
  printf("%d\n", d[n][0]);
}
