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 inf = 1061109567;
char s[1005];
int n;
int mp[5][1010], w[6], mat[6], f[1005][1 << 12], bin[5][5];
void work() {
  int s0 = 0, lim = 1;
  for (int j = 1; j <= 3; ++j)
    for (int i = 1; i <= 4; ++i) bin[i][j] = lim, lim <<= 1;
  for (int i = 1; i <= 4; ++i)
    for (int j = 1; j <= 3; ++j)
      if (mp[i][j]) s0 |= bin[i][j];
  f[1][s0] = 0;
  for (int j = 1; j <= n; ++j) {
    int now = 0;
    for (int i = 1; i <= 4; ++i)
      if (mp[i][j + 3]) now |= bin[i][1];
    for (int zt = 0; zt < lim; ++zt) {
      if (f[j][zt] == inf) continue;
      f[j + 1][lim - 1] = min(f[j + 1][lim - 1], f[j][zt] + w[4]);
      for (int a = 0; a <= 3; ++a)
        for (int b = 0; b <= 3; ++b)
          for (int c = 0; c <= 2; ++c)
            for (int d = 0; d <= 1; ++d) {
              int fy = w[a] + w[b] + w[c] + w[d];
              int kl = mat[a] | (mat[b] << 1) | (mat[c] << 2) | (mat[d] << 3);
              if (((zt | kl) & 15) == 15) {
                int nxt = ((zt | kl) >> 4) | (now << 8);
                f[j + 1][nxt] = min(f[j + 1][nxt], f[j][zt] + fy);
              }
            }
    }
  }
  printf("%d\n", f[n + 1][lim - 1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; ++i) scanf("%d", &w[i]);
  for (int i = 0; i < 4; ++i) {
    scanf("%s", s);
    for (int j = 0; j < n; ++j) mp[i + 1][j + 1] = (s[j] == '.');
  }
  for (int j = n + 1; j <= n + 3; ++j)
    for (int i = 1; i <= 4; ++i) mp[i][j] = 1;
  mat[1] = 1, mat[2] = 51, mat[3] = 1911;
  memset(f, 0x3f, sizeof(f));
  work();
  return 0;
}
