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;
template <typename T>
void read_to_vector(size_t N, std::vector<T> &v) {
  for (size_t i = 0; i < N; i++) {
    T tmp;
    std::cin >> tmp;
    v.push_back(tmp);
  }
}
template <typename T>
void print_vector(const std::vector<T> &v) {
  for (auto it = v.begin(); it < v.end(); it++) {
    if (it != v.begin()) std::cout << " ";
    std::cout << (*it);
  }
}
long long n;
vector<long long> a;
vector<int> cols;
const long long INF = 1000000000000000000;
const int fills[9] = {1 << 8,        1 << 9,        1 << 10,
                      1 << 11,       0b110011 << 4, 0b110011 << 5,
                      0b110011 << 6, 0b11101110111, 0b11101110111 << 1};
void solve() {
  cin >> n;
  read_to_vector(4, a);
  vector<string> ss;
  read_to_vector(4, ss);
  cols = vector<int>(n + 9, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 4; j++) {
      if (ss[j][i] == '*') {
        cols[i] |= 1 << j;
      }
    }
  }
  long long fillcost[9] = {a[0], a[0], a[0], a[0], a[1],
                           a[1], a[1], a[2], a[2]};
  vector<vector<long long> > maxcost(n + 9, vector<long long>(1 << 12, INF));
  maxcost[0][cols[0] << 8 | cols[1] << 4 | cols[2]] = 0;
  for (int i = 0; i < n; i++) {
    for (int mask = (1 << 12) - 1; mask >= 0; mask--) {
      if (maxcost[i][mask] == INF) {
        continue;
      }
      for (int filli = 0; filli < 9; filli++) {
        int x = ~fills[filli];
        int newmask =
            mask & x & (cols[i] << 8 | cols[i + 1] << 4 | cols[i + 2]);
        maxcost[i][newmask] =
            min(maxcost[i][newmask], maxcost[i][mask] + fillcost[filli]);
        if ((newmask & (0b1111 << 8)) == 0) {
          int nextmask = (newmask << 4 & ((1 << 12) - 1)) | cols[i + 3];
          maxcost[i + 1][nextmask] =
              min(maxcost[i + 1][nextmask], maxcost[i][newmask]);
        }
      }
      for (int j = i + 1; j <= i + 4; j++) {
        maxcost[j][cols[j] << 8 | cols[j + 1] << 4 | cols[j + 2] << 0] =
            min(maxcost[j][cols[j] << 8 | cols[j + 1] << 4 | cols[j + 2] << 0],
                maxcost[i][mask] + a[3]);
      }
    }
  }
  long long res = maxcost[n][0];
  cout << res << endl;
}
void pre_compute() {}
int main() {
  std::cout << std::setprecision(15);
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  pre_compute();
  solve();
  return 0;
}
