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 MAXN = 1e3 + 5;
const int MAXM = 4 + 5;
const int ALL = (1 << 8) + 5;
const int inf = 0x3f3f3f3f;
inline void chk_min(int &a, int b) {
  if (a > b) a = b;
}
int a[MAXM];
char s[MAXN][MAXM];
int f[MAXN][MAXM][ALL], to[ALL][MAXM][MAXM];
bool fob[ALL];
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= 4; ++i) scanf("%d", &a[i]);
  for (int j = 1; j <= 4; ++j) {
    static char ss[MAXN];
    scanf("%s", ss + 1);
    for (int i = 1; i <= n; ++i) s[i][j] = ss[i];
  }
  int all = (1 << 8) - 1;
  for (int mask = 0; mask <= all; ++mask) {
    for (int i = 0; i < 4; ++i)
      if ((((mask) >> ((i) << 1)) & 3) == 3) fob[mask] = 1;
    for (int i = 1; i <= 4; ++i)
      for (int j = 0; j <= i; ++j) {
        int &cur = to[mask][i][j] = mask;
        for (int k = i - j + 1; k <= i; ++k)
          if ((((cur) >> ((k - 1) << 1)) & 3) <= j)
            cur &= ~(3 << ((k - 1) << 1));
      }
  }
  memset(f, 0x3f, sizeof(f));
  f[0][4][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int mask = 0; mask <= all; ++mask)
      if (f[i - 1][4][mask] < inf) {
        f[i][0][mask] = f[i - 1][4][mask];
        if (fob[mask])
          chk_min(f[i][4][0], f[i][0][mask] + a[4]), f[i][0][mask] = inf;
      }
    for (int j = 1; j <= 4; ++j)
      for (int mask = 0; mask <= all; ++mask)
        if (f[i][j - 1][mask] < inf) {
          int nmask = mask;
          if ((((mask) >> ((j - 1) << 1)) & 3) || s[i][j] == '*')
            nmask += (1 << ((j - 1) << 1));
          for (int k = 0; k <= j && k <= i; ++k)
            chk_min(f[i][j][to[nmask][j][k]], f[i][j - 1][mask] + a[k]);
        }
  }
  printf("%d", f[n][4][0]);
  return 0;
}
