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 = 1e3 + 5;
const long long INF = 1e16 + 7;
char ff[5][N];
int cost[5];
long long DP[N][5][1 << 8];
long long fun(int n, int p, int fill) {
  if (DP[n][p][fill] != -1) return DP[n][p][fill];
  if (n == 0) {
    if (fill)
      return DP[n][p][fill] = INF;
    else
      return DP[n][p][fill] = 0;
  }
  if (p == 0) return DP[n][p][fill] = fun(n - 1, 4, fill);
  bitset<8> bs(fill);
  long long ans = INF;
  int f = (p << 1) - 2;
  int s = f + 1;
  int msk = (bs[f] << 1) + bs[s];
  if (ff[p][n] == '.' || bs[f] || bs[s]) {
    int tmsk = max(0, msk - 1);
    bitset<8> ts = bs;
    ts[f] = tmsk >> 1;
    ts[s] = tmsk % 2;
    ans = min(ans, fun(n, p - 1, ts.to_ulong()));
  }
  for (int j = 1; j <= p; j++) {
    bitset<8> ts = bs;
    int ff = f, ss = s;
    for (int k = p; k > p - j; k--) {
      int msk = (bs[ff] << 1) + bs[ss];
      int tmsk = max(msk, j);
      if (j == p || k == p) tmsk--;
      ts[ff] = tmsk >> 1;
      ts[ss] = tmsk % 2;
      ff -= 2;
      ss = ff + 1;
    }
    if (j == p)
      ans = min(ans, cost[j] + fun(n, p - j, ts.to_ulong()));
    else
      ans = min(ans, cost[j] + fun(n, p - 1, ts.to_ulong()));
  }
  return DP[n][p][fill] = ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) {
    scanf("%d", &cost[i]);
  }
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      scanf(" %c", &ff[i][j]);
    }
  }
  memset(DP, -1, sizeof DP);
  printf("%lld\n", fun(n, 4, 0));
  return 0;
}
