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 out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 1111;
const int inf = 2e9;
int n, a1, a2, a3, a4;
string g[4];
int dp[1 << 12][maxn];
int ans = inf;
int get(int col) {
  int mask = 0;
  for (int i = 0; i < 3 && col + i < n; i++) {
    for (int j = 0; j < 4; j++) {
      if (g[j][col + i] == '.') {
        mask = mask | (1 << (4 * i + j));
      }
    }
  }
  return mask;
}
void solve(int i, int mask) {
  if (i == n) {
    ans = min(ans, dp[mask][i]);
  }
  if (dp[mask][i] == inf) return;
  if ((mask & 15) == 15) {
    int mask2 = (mask / 16) | get(i + 1);
    dp[mask2][i + 1] = min(dp[mask2][i + 1], dp[mask][i]);
  }
  for (int j = 0; j < 4; j++) {
    int mask2 = mask | (1 << j);
    dp[mask2][i] = min(dp[mask2][i], dp[mask][i] + a1);
  }
  for (int j = 0; j < 3; j++) {
    int mask2 = mask;
    for (int x = 0; x < 2; x++) {
      for (int y = 0; y < 2; y++) {
        mask2 = mask2 | (1 << (j + 4 * x + y));
      }
    }
    dp[mask2][i] = min(dp[mask2][i], dp[mask][i] + a2);
  }
  for (int j = 0; j < 2; j++) {
    int mask2 = mask;
    for (int x = 0; x < 3; x++) {
      for (int y = 0; y < 3; y++) {
        mask2 = mask2 | (1 << (j + 4 * x + y));
      }
    }
    dp[mask2][i] = min(dp[mask2][i], dp[mask][i] + a3);
  }
  if (i + 4 >= n) {
    ans = min(ans, dp[mask][i] + a4);
  } else {
    dp[get(i + 4)][i + 4] = min(dp[get(i + 4)][i + 4], dp[mask][i] + a4);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a1 >> a2 >> a3 >> a4;
  for (int i = 0; i < 4; i++) {
    cin >> g[i];
  }
  for (int i = 0; i < maxn; i++) {
    for (int mask = 0; mask < (1 << 12); mask++) {
      dp[mask][i] = inf;
    }
  }
  dp[get(0)][0] = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < (1 << 12); j++) {
      solve(i, j);
    }
  }
  cout << ans << endl;
  return 0;
}
