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;
int cost[5];
bool inp[5][1005];
int mat[1005];
vector<int> op[5];
int memo[1005][4096];
int solve(int i, int st) {
  if (i > n) return 0;
  if (memo[i][st] != -1) return memo[i][st];
  int mn = 1000000000;
  if (i != n) {
    if (st & (15 << 8))
      mn = cost[4] + solve(i + 2, mat[i + 2]);
    else
      mn = solve(i + 1, (st << 4) | mat[i + 1]);
  } else if (i == n && !st)
    mn = 0;
  for (int j = (1); j <= (4); j++) {
    for (auto o : op[j]) {
      if ((st & o) == st) continue;
      int nst = st & o;
      mn = min(mn, cost[j] + solve(i, nst));
      if (i == n) {
        if (!nst) mn = min(mn, cost[j]);
      } else {
        if (nst & (15 << 8)) {
          mn = min(mn, cost[j] + cost[4] + solve(i + 2, mat[i + 2]));
        } else {
          mn = min(mn, cost[j] + solve(i + 1, (nst << 4) | mat[i + 1]));
        }
      }
    }
  }
  return memo[i][st] = mn;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (4); i++) scanf("%d", &cost[i]);
  for (int i = (1); i <= (4); i++) {
    getchar();
    int t = 0;
    for (int j = (1); j <= (n); j++) {
      char c = getchar();
      inp[i][j] = (c == '*');
    }
  }
  for (int j = (1); j <= (n); j++) {
    int t = 0;
    for (int i = (1); i <= (4); i++) t <<= 1, t |= inp[i][j];
    mat[j] = t;
  }
  for (int i = (1); i <= (4); i++) {
    for (int j = (1); j <= (5 - i); j++) {
      for (int k = (1); k <= (5 - i); k++) {
        int t = 0;
        for (int l = (j); l <= (j + i - 1); l++) {
          for (int m = (k); m <= (k + i - 1); m++) {
            t |= (1 << ((4 - m) * 4 + 4 - l));
          }
        }
        t ^= 65535;
        op[i].push_back(t);
      }
    }
  }
  memset(memo, -1, sizeof(memo));
  printf("%d\n", solve(1, mat[1]));
  return 0;
}
