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 N = 1005, inf = 0x3f3f3f3f, B = ((1 << 16) - 1);
int n, c[5], f[2][B + 5], zz[N];
char mp[5][N];
vector<int> ve[5];
int main() {
  int i, j, k, l, t, z, lz, up, tmp, o, S;
  scanf("%d", &n);
  for (i = 1; i <= 4; i++) scanf("%d", &c[i]);
  for (i = 0; i < 4; i++) scanf("%s", mp[i]);
  for (i = 0; i < n; i++) {
    for (j = 0; j < 4; j++) zz[i] = (zz[i] << 1) + (mp[j][i] != '.') ? 1 : 0;
  }
  for (i = 1; i <= 4; i++) {
    for (j = 0; j + i <= 4; j++) {
      o = B;
      for (k = j; k < j + i; k++)
        for (l = 0; l < i; l++) o ^= (1 << (l * 4 + k));
      ve[i].push_back(o);
    }
  }
  for (i = 0; i < 2; i++)
    for (j = B; ~j; j--) f[i][j] = inf;
  for (i = 0; i < 4; i++)
    for (j = 0; j < 4; j++)
      if (mp[i][j] == '*') S |= (1 << (j * 4 + i));
  f[0][S] = 0;
  for (i = 0; i < n; i++) {
    t = i & 1;
    for (lz = B; ~lz; lz--)
      if (f[t][lz] != inf) {
        if (!(lz & 15)) {
          z = (lz >> 4);
          for (j = 0; j < 4; j++)
            if (mp[j][i + 4] == '*') z |= (1 << (12 + j));
          f[t ^ 1][z] = f[t][lz];
        }
        for (j = 1; j <= 4; j++) {
          for (vector<int>::iterator it = ve[j].begin(); it != ve[j].end();
               it++)
            f[t][lz & (*it)] = min(f[t][lz & (*it)], f[t][lz] + c[j]);
        }
      }
    memset(f[t], inf, sizeof f[t]);
  }
  printf("%d\n", f[n & 1][0]);
}
