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, now, pos;
char c;
const int N = 20, M = 1005;
int w[N], f[M], mx[N][N], m[N][N], dp[M][(1 << 16) + 5];
int main() {
  cin >> n;
  for (int i = 1; i <= 4; ++i) scanf("%d", &w[i]);
  for (int i = 0; i < 4; ++i)
    for (int j = 0; j < n; ++j) {
      cin >> c;
      f[j] <<= 1;
      if (c == '*') ++f[j];
    }
  for (int i = 0; i < 4; ++i)
    for (int j = 1; j <= 4 - i; ++j) {
      for (int k = 0; k < 4; ++k)
        for (int p = 0; p < 4; ++p) m[k][p] = 1;
      for (int k = 0; k < j; ++k)
        for (int p = 0; p < j; ++p) m[k][i + p] = 0;
      now = 0;
      for (int k = 3; k >= 0; --k)
        for (int p = 0; p < 4; ++p) now = (now << 1) + m[k][p];
      mx[i][j] = now;
    }
  memset(dp, 0x3f3f3f3f, sizeof(dp));
  for (int i = 3; i >= 0; --i) pos <<= 4, pos += f[i];
  dp[0][pos] = 0;
  for (int i = 0; i <= n; ++i)
    for (int j = (1 << 16) - 1; j >= 0; --j) {
      if (dp[i][j] >= 0x3f3f3f3f - 10) continue;
      if (!(j & 15)) {
        int now = (f[i + 4] << 12) | (j >> 4);
        dp[i + 1][now] = min(dp[i + 1][now], dp[i][j]);
      }
      for (int k = 0; k < 4; ++k)
        for (int p = 1; p <= 4 - k; ++p)
          dp[i][j & mx[k][p]] = min(dp[i][j & mx[k][p]], dp[i][j] + w[p]);
    }
  cout << dp[n][0];
  return 0;
}
