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 n, arr[5], dp[4][1005][5][5][5][5];
char s[4][1005];
int solve(int i = 0, int j = 0, int a = 0, int b = 0, int c = 0, int d = 0) {
  if (i == 4)
    return solve(0, j + 1, max(a - 1, 0), max(b - 1, 0), max(c - 1, 0),
                 max(d - 1, 0));
  if (j == n) return 0;
  int &ret = dp[i][j][a][b][c][d];
  if (~ret) return ret;
  ret = 1e9;
  int A[] = {a, b, c, d}, B[] = {a, b, c, d};
  if (A[i] || s[i][j] == '.') ret = solve(i + 1, j, a, b, c, d);
  for (int x = 1; i + x <= 4 && j + x <= n; x++) {
    for (int pos = i; pos < i + x; pos++) A[pos] = max(A[pos], x);
    ret = min(ret, solve(i + 1, j, A[0], A[1], A[2], A[3]) + arr[x]);
    for (int pos = i; pos < i + x; pos++) A[pos] = B[pos];
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", arr + i);
  for (int i = 0; i < 4; i++) scanf("%s", s[i]);
  memset(dp, -1, sizeof dp);
  printf("%d", solve());
  return 0;
}
