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;
const long long inf = 1e18;
const long long N = 1e3 + 5;
const long long MaxM = (1 << 12) + 5;
long long n, f[N + 10][MaxM], a[10][N + 10], M, c1, c2, c3, c4, res = 1e18;
long long get(long long c) {
  long long res = 0;
  for (long long i = 0; i <= 2; i++) {
    if (c + i < n) {
      for (long long j = 0; j <= 3; j++) {
        if (a[j][c + i] == 1) res ^= (1 << (j + 4 * i));
      }
    }
  }
  return res;
}
void Solve(long long col, long long mask) {
  if (f[col][mask] == inf) return;
  if (col == n) {
    res = min(res, f[col][mask]);
    return;
  }
  if ((mask & 15) == 15) {
    f[col + 1][(mask / 16) | get(col + 1)] =
        min(f[col + 1][(mask / 16) | get(col + 1)], f[col][mask]);
    return;
  }
  for (long long i = 0; i <= 3; i++) {
    long long CUR = mask | (1 << i);
    f[col][CUR] = min(f[col][CUR], f[col][mask] + c1);
  }
  for (long long i = 0; i <= 2; i++) {
    long long CUR = mask;
    for (long long j = 0; j <= 1; j++)
      for (long long k = 0; k <= 1; k++) {
        CUR |= (1 << (i + j + 4 * k));
      }
    f[col][CUR] = min(f[col][CUR], f[col][mask] + c2);
  }
  for (long long i = 0; i <= 1; i++) {
    long long CUR = mask;
    for (long long j = 0; j <= 2; j++)
      for (long long k = 0; k <= 2; k++) {
        CUR |= (1 << (i + j + 4 * k));
      }
    f[col][CUR] = min(f[col][CUR], f[col][mask] + c3);
  }
  if (col + 4 >= n) {
    res = min(res, f[col][mask] + c4);
    return;
  }
  f[col + 4][get(col + 4)] = min(f[col + 4][get(col + 4)], f[col][mask] + c4);
}
void in() {
  cin >> n;
  cin >> c1 >> c2 >> c3 >> c4;
  for (long long i = 0; i <= 3; i++)
    for (long long j = 0; j <= n - 1; j++) {
      char c;
      cin >> c;
      if (c == '.')
        a[i][j] = 1;
      else
        a[i][j] = 0;
    }
  M = (1 << 12) - 1;
  for (long long i = 0; i <= n + 5; i++)
    for (long long mask = 0; mask <= M; mask++) f[i][mask] = inf;
}
void process() {
  f[0][get(0)] = 0;
  for (long long i = 0; i <= n; i++)
    for (long long mask = 0; mask <= M; mask++) {
      Solve(i, mask);
    }
  cout << res << '\n';
}
int main() {
  if (fopen("trash.inp", "r"))
    freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  in();
  process();
}
