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 const N = 1000 + 10;
int const M = (1 << 16) + 10;
int const inf = 0x7f7f7f7f;
int a[5], n;
char mp[5][N];
int dp[2][M];
vector<int> p[5];
void Init() {
  for (int i = 1; i <= 4; i++) {
    for (int j = 0; j + i <= 4; j++) {
      int v = (1 << 16) - 1;
      for (int r = j; r < j + i; r++)
        for (int k = 0; k < i; k++) v ^= (1 << (4 * k + r));
      p[i].push_back(v);
    }
  }
}
void solve() {
  int state = 0;
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < 4; j++)
      if (mp[i][j] == '*') state |= 1 << (4 * j + i);
  memset(dp, inf, sizeof(dp));
  int up = (1 << 0) + (1 << 1) + (1 << 2) + (1 << 3);
  dp[0][state] = 0;
  for (int i = 0; i < n; i++) {
    int id = i & 1;
    for (int j = (1 << 16) - 1; j >= 0; j--) {
      if (dp[id][j] == inf) continue;
      if (!(j & up)) {
        int tmp = j >> 4;
        for (int k = 0; k < 4; k++)
          if (mp[k][i + 4] == '*') tmp |= 1 << (4 * 3 + k);
        dp[id ^ 1][tmp] = dp[id][j];
      }
      for (int k = 1; k <= 4; k++) {
        for (int l = 0; l < p[k].size(); l++)
          dp[id][j & p[k][l]] = min(dp[id][j & p[k][l]], dp[id][j] + a[k]);
      }
    }
    memset(dp[id], inf, sizeof(dp[id]));
  }
  printf("%d\n", dp[n & 1][0]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
  for (int i = 0; i < 4; i++) scanf(" %s", mp[i]);
  Init();
  solve();
  return 0;
}
