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 = 1010;
int n, cost[5], g[5], f[N][1 << 12];
char str[5][N];
inline void up(int& x, int y) {
  if (x < 0 || y < x) x = y;
}
int main() {
  int i, j, k, a, b, c, d, A;
  scanf("%d", &n);
  for (i = 1; i <= 4; ++i) scanf("%d", &cost[i]);
  for (i = 1; i <= 4; ++i) scanf("%s", str[i] + 1);
  g[1] = 1, g[2] = 3 * (1 + 16), g[3] = 7 * (1 + 16 + 256),
  g[4] = (1 << 16) - 1;
  int s0 = 0;
  for (i = 1; i <= 4; ++i)
    for (j = 1; j <= 3; ++j)
      if (str[i][j] == '.') s0 |= 1 << ((j - 1) * 4 + i - 1);
  memset(f, 255, sizeof(f));
  f[0][s0] = 0;
  for (i = 0; i < n; ++i) {
    int sr = 0;
    for (j = 1; j <= 4; ++j)
      if (str[j][i + 4] == '.') sr |= 1 << (j - 1);
    for (A = 0; A < (1 << 12); ++A)
      if (~f[i][A]) {
        for (d = 0; d <= 4; ++d)
          for (c = 0; c <= 3; ++c)
            for (b = 0; b <= 2; ++b)
              for (a = 0; a <= 1; ++a) {
                int sumc = cost[a] + cost[b] + cost[c] + cost[d];
                int s0 = g[a] << (1 - a) | g[b] << (2 - b) | g[c] << (3 - c) |
                         g[d] << (4 - d);
                if ((A & 15 | s0 & 15) == 15)
                  up(f[i + 1][A >> 4 | s0 >> 4 | sr << 8], f[i][A] + sumc);
              }
      }
  }
  printf("%d", f[n][0]);
  return 0;
}
