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 N = 1 << 10, M = 1 << 4;
int n, x[N];
char s[4][N];
long long mem[N][M][M][M];
long long a[5];
long long dp(int i, int a1, int a2, int a3) {
  if (i >= n) return 0ll;
  long long &ret = mem[i][a1][a2][a3];
  if (ret != -1) return ret;
  if (a1 == 0) return ret = dp(i + 1, a2, a3, x[i + 3]);
  ret = a[4] + dp(i + 4, x[i + 4], x[i + 5], x[i + 6]);
  ret = min(ret, a[3] + dp(i, a1 & 8, a2 & 8, a3 & 8));
  ret = min(ret, a[3] + dp(i, a1 & 1, a2 & 1, a3 & 1));
  ret = min(ret, a[2] + dp(i, a1 & 12, a2 & 12, a3));
  ret = min(ret, a[2] + dp(i, a1 & 9, a2 & 9, a3));
  ret = min(ret, a[2] + dp(i, a1 & 3, a2 & 3, a3));
  ret = min(ret, a[1] + dp(i, a1 & 14, a2, a3));
  ret = min(ret, a[1] + dp(i, a1 & 13, a2, a3));
  ret = min(ret, a[1] + dp(i, a1 & 11, a2, a3));
  ret = min(ret, a[1] + dp(i, a1 & 7, a2, a3));
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%lld", a + i);
  for (int i = 0; i < 4; i++) scanf("%s", s[i]);
  for (int j = 0; j < 4; j++)
    for (int i = 0; i < n; i++) x[i] += ((s[j][i] == '*') << j);
  n += 5;
  memset(mem, -1, sizeof mem);
  printf("%lld\n", dp(0, x[0], x[1], x[2]));
  return 0;
}
