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;
int n, pd[1010][5 * 5 * 5 * 5], c[4];
char s[4][1010];
int solve(int i, int mask) {
  int& ret = pd[i][mask];
  if (ret != -1) return ret;
  if (i == n) return ret = 0;
  ret = 3e8;
  int v[4], tmask = mask, ok = 1;
  for (__typeof(0) j = (4) - ((4) > (0)); j != (0) - ((4) > (0));
       j += 1 - 2 * ((4) > (0))) {
    v[j] = tmask % 5;
    tmask /= 5;
    if (!v[j] && s[j][i] == '*') ok = 0;
  }
  if (ok) {
    tmask = 0;
    for (__typeof(4) j = (0) - ((0) > (4)); j != (4) - ((0) > (4));
         j += 1 - 2 * ((0) > (4)))
      tmask = 5 * tmask + max(v[j] - 1, 0);
    ret = min(ret, solve(i + 1, tmask));
  }
  int pv[4];
  for (__typeof(4) j = (0) - ((0) > (4)); j != (4) - ((0) > (4));
       j += 1 - 2 * ((0) > (4)))
    for (__typeof(5 - j) k = (1) - ((1) > (5 - j));
         k != (5 - j) - ((1) > (5 - j)); k += 1 - 2 * ((1) > (5 - j)))
      if (i + k <= n) {
        int ok = 0;
        for (__typeof(4) l = (0) - ((0) > (4)); l != (4) - ((0) > (4));
             l += 1 - 2 * ((0) > (4)))
          pv[l] = v[l];
        for (__typeof(j + k) l = (j) - ((j) > (j + k));
             l != (j + k) - ((j) > (j + k)); l += 1 - 2 * ((j) > (j + k)))
          if (v[l] < k) {
            pv[l] = k;
            ok = 1;
          }
        if (!ok) continue;
        tmask = 0;
        for (__typeof(4) l = (0) - ((0) > (4)); l != (4) - ((0) > (4));
             l += 1 - 2 * ((0) > (4)))
          tmask = 5 * tmask + pv[l];
        ret = min(ret, solve(i, tmask) + c[k - 1]);
      }
  return ret;
}
int main() {
  memset(pd, -1, sizeof(pd));
  scanf("%d", &n);
  for (__typeof(4) i = (0) - ((0) > (4)); i != (4) - ((0) > (4));
       i += 1 - 2 * ((0) > (4)))
    scanf("%d", c + i);
  for (__typeof(4) i = (0) - ((0) > (4)); i != (4) - ((0) > (4));
       i += 1 - 2 * ((0) > (4)))
    scanf("%s", s[i]);
  printf("%d\n", solve(0, 0));
}
