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, dp[5][1010][(1 << 12)], p[4];
bool bio[5][1010][(1 << 12)];
string s[5];
int rek(int x, int y, int mask) {
  if (y >= n) return 0;
  if (bio[x][y][mask]) return dp[x][y][mask];
  bio[x][y][mask] = 1;
  int ret = 1e9;
  for (int i = 1; i <= 3; i++) {
    if (x - i >= -1) {
      int newmask = 0;
      for (int j = 0; j < 3; j++) {
        for (int k = 0; k <= 3; k++) {
          if (k >= i || j >= i) {
            newmask += (mask & (1 << (j * 4 + k)));
          }
        }
      }
      newmask /= 2;
      if (s[x][y + 3] == '*') newmask += (1 << 11);
      int newx, newy;
      if (x == 0) {
        newx = 3;
      } else
        newx = x - 1;
      if (newx == 3)
        newy = y + 1;
      else
        newy = y;
      ret = min(ret, rek(newx, newy, newmask) + p[i]);
    }
  }
  if (x == 3) {
    int newmask = 0;
    for (int i = y + 4; i < y + 7; i++) {
      for (int j = 3; j >= 0; j--) {
        if (s[j][i] == '*') newmask += (1 << ((i - y - 4) * 4 + 3 - j));
      }
    }
    ret = min(ret, rek(x, y + 4, newmask) + p[4]);
  }
  if ((mask & 1) == 0) {
    int newmask = mask / 2;
    if (s[x][y + 3] == '*') newmask += (1 << 11);
    if (x > 0) {
      ret = min(ret, rek(x - 1, y, newmask));
    } else {
      ret = min(ret, rek(3, y + 1, newmask));
    }
  }
  return dp[x][y][mask] = ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= 4; i++) cin >> p[i];
  for (int i = 0; i < 4; i++) {
    cin >> s[i];
    s[i] += '.';
    s[i] += '.';
    s[i] += '.';
    s[i] += '.';
    s[i] += '.';
    s[i] += '.';
  }
  int maska = 0;
  for (int i = 0; i <= 2; i++) {
    for (int j = 3; j >= 0; j--) {
      if (s[j][i] == '*') maska += (1 << (i * 4 + 3 - j));
    }
  }
  cout << rek(3, 0, maska);
}
