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, c1, c2, c3, c4, a[5][1010], F[1010][4100];
bool Free[1010][4100];
int ID(int i, int j) { return (i - 1) * 3 + j - 1; }
int GetBit(int x, int k) { return (x >> k) & 1; }
int Del(int mask, int u1, int v1, int u2, int v2) {
  for (int i = u1; i <= u2; i++)
    for (int j = v1; j <= v2; j++)
      if (GetBit(mask, 3 * i + j) == 1) mask -= (1 << (3 * i + j));
  return mask;
}
int f(int i, int mask) {
  if (i == 3 && mask == 0) int h = 0;
  if (Free[i][mask] == true) return F[i][mask];
  Free[i][mask] = true;
  if (i > n)
    F[i][mask] = 0;
  else if (GetBit(mask, 0) + GetBit(mask, 3) + GetBit(mask, 6) +
               GetBit(mask, 9) ==
           0) {
    int newmask = mask >> 1;
    newmask += a[1][i + 3] * (1 << 2) + a[2][i + 3] * (1 << 5) +
               a[3][i + 3] * (1 << 8) + a[4][i + 3] * (1 << 11);
    F[i][mask] = f(i + 1, newmask);
  } else {
    F[i][mask] = 1e9;
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 0, 0, 0, 0)) + c1);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 1, 0, 1, 0)) + c1);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 2, 0, 2, 0)) + c1);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 3, 0, 3, 0)) + c1);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 0, 0, 1, 1)) + c2);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 1, 0, 2, 1)) + c2);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 2, 0, 3, 1)) + c2);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 0, 0, 2, 2)) + c3);
    F[i][mask] = min(F[i][mask], f(i, Del(mask, 1, 0, 3, 2)) + c3);
    int newmask = a[1][i + 4] * (1 << 2) + a[2][i + 4] * (1 << 5) +
                  a[3][i + 4] * (1 << 8) + a[4][i + 4] * (1 << 11);
    F[i][mask] = min(F[i][mask], f(i + 2, newmask) + c4);
  }
  return F[i][mask];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> c1 >> c2 >> c3 >> c4;
  char c;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= n; j++) {
      cin >> c;
      if (c == '*') a[i][j] = 1;
    }
  int mask = 0;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= 3; j++) mask += a[i][j] * (1 << ID(i, j));
  cout << f(1, mask);
}
