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 M = 1e9 + 7, N = 1e3 + 5;
int dp[4][N][1 << 12], cost[4], add[4][4], n;
bool is[5][N];
int calc(int c, int r, int msk) {
  if (r == 4) return !(msk & 15) ? calc(c + 1, 0, (msk >> 4) | (15 << 8)) : 1e9;
  if (c >= n) return 0;
  int &ref = dp[r][c][msk];
  if (ref != -1) return ref;
  if (!is[r][c]) msk &= ~(1 << r);
  ref = calc(c, r + 1, msk);
  for (int i = 0; i < 4; i++)
    if (i < 3)
      ref = min(ref, calc(c, r + 1, (msk & ~add[r][i])) + cost[i]);
    else
      ref = min(ref, calc(c + 4, 0, (1 << 12) - 1) + cost[i]);
  return ref;
}
int main() {
  scanf("%d", &n);
  memset(dp, -1, sizeof dp);
  for (int i = 0; i < 4; i++) scanf("%d", cost + i);
  for (int i = 0; i < 3; i++)
    for (int j = 0; j + i < 4; j++)
      for (int k = 0; k <= i; k++)
        for (int l = 0; l <= i; l++) add[j][i] |= 1 << (4 * k + j + l);
  for (int i = 0; i < 4; i++) {
    char c[N];
    scanf("%s", c);
    string s = c;
    for (int j = 0; j < s.size(); j++) is[i][j] = s[j] != '.';
  }
  printf("%d\n", calc(0, 0, (1 << 12) - 1));
}
