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 = 1015;
const int INF = 1e9;
int n, a[4];
char s[4][N];
map<pair<int, int>, int> dp;
int tmask[5] = {0, 1, 0x33, 0x777, 0xffff};
inline bool intersect(int id, int mask) {
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      if ((mask & (1 << j)) && s[j][id + i] == '*') return true;
    }
    mask >>= 4;
  }
  return false;
}
int cal(int mask, int id) {
  if (id >= n) return 0;
  auto it = dp.find({mask, id});
  if (it != dp.end()) return it->second;
  int val = INF;
  for (int k = 1; k <= 4; k++) {
    int mk = tmask[k];
    for (int i = 0; i <= 4 - k; i++, mk <<= 1) {
      if (mk & mask) {
        int nmask = mask ^ (mk & mask);
        bool ist = intersect(id, mk & mask);
        if (nmask & 0xf)
          val = min(val, a[k - 1] * ist + cal(nmask, id));
        else if (nmask & 0xf0)
          val = min(val, a[k - 1] * ist + cal((nmask >> 4) | 0xf000, id + 1));
        else if (nmask & 0xf00)
          val = min(val, a[k - 1] * ist + cal((nmask >> 8) | 0xff00, id + 2));
        else if (nmask & 0xf000)
          val = min(val, a[k - 1] * ist + cal((nmask >> 12) | 0xfff0, id + 3));
        else
          val = min(val, a[k - 1] * ist + cal(0xffff, id + 4));
      }
    }
  }
  return dp[{mask, id}] = val;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 4; i++) {
    scanf("%d", a + i);
  }
  for (int i = 0; i < 4; i++) {
    scanf("%s", s[i]);
  }
  printf("%d\n", cal(0xffff, 0));
  return 0;
}
