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 C[1 << 16];
int inf = 1e8;
int n;
int Mask[1005];
vector<int> M;
string str[4];
int dp[1 << 12][1005];
int call(int lastMask, int col) {
  if (col == n) return 0;
  if (dp[lastMask][col] != -1) return dp[lastMask][col];
  int ans = inf;
  int curMask = lastMask + (Mask[col] << 12);
  for (int m : M) {
    if (((m | curMask) & 15) == 15) {
      int nxtMask = ((m | curMask) >> 4);
      ans = min(ans, C[m] + call(nxtMask, col + 1));
    }
  }
  return dp[lastMask][col] = ans;
}
int main() {
  cin >> n;
  int a[5];
  for (int i = 1; i <= 4; i++) {
    cin >> a[i];
  }
  vector<int> moves = {1, 2, 4, 8, 51, 102, 204, 1911, 3822, 65535};
  vector<int> cst = {a[1], a[1], a[1], a[1], a[2],
                     a[2], a[2], a[3], a[3], a[4]};
  for (int i = 0; i < (1 << 16); i++) {
    C[i] = inf;
  }
  for (int mask = 0; mask < (1 << 10); mask++) {
    int m = 0, c = 0;
    for (int k = 0; k < 10; k++) {
      if ((mask >> k) & 1) {
        m |= moves[k];
        c += cst[k];
      }
    }
    C[m] = min(C[m], c);
  }
  for (int i = 0; i < (1 << 16); i++) {
    if (C[i] != inf) M.push_back(i);
  }
  for (int i = 0; i < 4; i++) {
    cin >> str[i];
  }
  for (int i = 0; i < n; i++) {
    int m = 0;
    for (int j = 0; j < 4; j++) {
      if (str[j][i] == '.') m |= (1 << j);
    }
    Mask[i] = m;
  }
  Mask[n] = Mask[n + 1] = Mask[n + 2] = 15;
  n += 3;
  memset(dp, -1, sizeof dp);
  int ans = call((1 << 12) - 1, 0);
  cout << ans << endl;
}
