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 lim = (1 << 4);
int n, w[4], g[1005], f[1002][16][16][16][16], s[4][4][4], num[4][4];
char str[4][1000];
int read() {
  int x = 0, f = 1;
  char s;
  while ((s = getchar()) > '9' || s < '0')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 1) + (x << 3) + (s ^ 48), s = getchar();
  return x * f;
}
int main() {
  int t;
  n = read();
  for (int i = 0; i < 4; ++i) w[i] = read();
  for (int i = 0; i < 4; ++i) scanf("%s", str[i]);
  for (int i = 0; i < n; ++i) {
    t = 0;
    for (int j = 0; j < 4; ++j) t <<= 1, t += (str[j][i] == '.');
    g[i] = t;
  }
  for (int i = n; i < n + 5; ++i) g[i] = lim - 1;
  memset(f, 0x3f, sizeof f);
  f[0][g[0]][g[1]][g[2]][g[3]] = 0;
  for (int i = 0; i < 4; ++i)
    for (int j = 0; j < 4 - i; ++j) {
      for (int k = 0; k < 4; ++k)
        for (int u = 0; u < 4; ++u) num[k][u] = 0;
      for (int k = 0; k <= i; ++k)
        for (int u = j; u <= j + i; ++u) num[k][u] = 1;
      for (int k = 0; k < 4; ++k) {
        t = 0;
        for (int u = 0; u < 4; ++u) t <<= 1, t += num[k][u];
        s[i][j][k] = t;
      }
    }
  for (int i = 0; i < n; ++i) {
    for (int a = 0; a < lim; ++a)
      for (int b = 0; b < lim; ++b)
        for (int c = 0; c < lim; ++c)
          for (int d = 0; d < lim; ++d) {
            if (f[i][a][b][c][d] >= 0x3f3f3f3f) continue;
            if (a == lim - 1)
              f[i + 1][b][c][d][g[i + 4]] =
                  min(f[i + 1][b][c][d][g[i + 4]], f[i][a][b][c][d]);
            for (int j = 0; j < 4; ++j)
              for (int k = 0; k < 4 - j; ++k) {
                f[i][a | s[j][k][0]][b | s[j][k][1]][c | s[j][k][2]]
                 [d | s[j][k][3]] = min(f[i][a | s[j][k][0]][b | s[j][k][1]]
                                         [c | s[j][k][2]][d | s[j][k][3]],
                                        f[i][a][b][c][d] + w[j]);
              }
          }
  }
  printf("%d\n", f[n - 1][lim - 1][lim - 1][lim - 1][lim - 1]);
  return 0;
}
