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 oo = 1e9;
int a[1010];
int c[5];
int dp[1005][1 << 4][1 << 4][1 << 4];
int n, m;
int getbit(int x, int i) { return (x >> i) & 1; }
void minimize(int &a, int b) {
  if (a > b) a = b;
}
void turn_off(int &x, int i) {
  if (getbit(x, i)) x -= (1 << i);
}
int solve(int cur, int fcol, int scol, int tcol) {
  int &res = dp[cur][fcol][scol][tcol];
  if (cur >= n + 4) return 0;
  if (!fcol) return solve(cur + 1, scol, tcol, a[cur]);
  if (res != -1) return res;
  res = solve(cur + 4, a[cur + 1], a[cur + 2], a[cur + 3]) + c[4];
  for (int sz = 1; sz <= 3; sz++)
    for (int i = 0; i <= 4 - sz; i++) {
      int nxt_fcol = fcol, nxt_scol = scol, nxt_tcol = tcol;
      for (int j = i; j <= i + sz - 1; j++) {
        turn_off(nxt_fcol, j);
        if (sz >= 2) turn_off(nxt_scol, j);
        if (sz >= 3) turn_off(nxt_tcol, j);
      }
      minimize(res, solve(cur, nxt_fcol, nxt_scol, nxt_tcol) + c[sz]);
    }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= 4; i++) {
    cin >> c[i];
  }
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= n; j++) {
      char c;
      cin >> c;
      a[j] += (c == '*') << (i - 1);
    }
  memset(dp, -1, sizeof(dp));
  cout << solve(4, a[1], a[2], a[3]);
  return 0;
}
