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 dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
bool check(int mask, int pos) { return mask & (1 << pos); }
int SET(int mask, int pos) { return mask | (1 << pos); }
int FLIP(int mask, int pos) { return mask ^ (1 << pos); }
int reset(int mask, int pos) { return mask & ~(1 << pos); }
int n, a[5];
string str[10];
int mask[1005];
int dp[1005][16][16][16][16];
int solve(int pos, int cur1, int cur2, int cur3, int cur4) {
  if (pos == n) {
    return 0;
  }
  int &ret = dp[pos][cur1][cur2][cur3][cur4];
  if (ret != -1) {
    return ret;
  }
  ret = 100000000;
  if (cur1 == 0) {
    return ret = solve(pos + 1, cur2, cur3, cur4, mask[pos + 4]);
  }
  for (int sagol = 0; sagol < 4; sagol++) {
    for (int sz = 1; sz <= 4; sz++) {
      if (sagol + sz > 4) break;
      int tmp[] = {cur1, cur2, cur3, cur4};
      for (int i = 0; i < sz; i++) {
        for (int j = 0; j < sz; j++) {
          tmp[i] = reset(tmp[i], j + sagol);
        }
      }
      if (cur1 != tmp[0] || cur2 != tmp[1] || cur3 != tmp[2] || cur4 != tmp[3])
        ret = min(ret, solve(pos, tmp[0], tmp[1], tmp[2], tmp[3]) + a[sz]);
    }
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 1; i <= 4; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < 4; i++) {
    cin >> str[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 4; j++) {
      if (str[j][i] == '*') {
        mask[i] = SET(mask[i], j);
      }
    }
  }
  memset(dp, -1, sizeof dp);
  cout << solve(0, mask[0], mask[1], mask[2], mask[3]) << '\n';
}
