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;
inline char readchar() {
  char c = getchar();
  while (c == ' ' || c == '\n' || c == '\r') c = getchar();
  return c;
}
int n;
int r[100100], a[5];
vector<int> mk[5];
inline void init() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", a + i);
  memset(r, 0, sizeof(r));
  for (int i = 1; i <= 4; i++)
    for (int j = 0; j < n; j++) {
      char c = readchar();
      if (c == '*') r[j] |= 1 << (i - 1);
    }
  for (int i = 1; i <= 4; i++)
    for (int j = 0; j + i <= 4; j++) {
      int x = 0;
      for (int I = 0; I < i; I++)
        for (int J = 0; J < i; J++) x |= 1 << (4 * J + j + I);
      mk[i].push_back(((1 << 16) - 1) ^ x);
    }
}
int f[2][70010];
int main() {
  init();
  memset(f, 0x3f, sizeof(f));
  int S = 0;
  for (int i = 0; i < min(4, n); i++) S |= (r[i] << (i << 2));
  f[0][S] = 0;
  for (int i = 0; i < n; i++) {
    int cur = i & 1, nxt = cur ^ 1;
    memset(f[nxt], 0x3f, sizeof(f[nxt]));
    for (int S = (1 << 16) - 1; S >= 0; S--) {
      if (f[cur][S] > 1e9) continue;
      if ((S & 15) == 0) {
        int SS = (S >> 4) | (r[i + 4] << 12);
        f[nxt][SS] = min(f[nxt][SS], f[cur][S]);
      }
      for (int j = 1; j <= 4 && i + j <= n; j++)
        for (int k = 0; k < (int)mk[j].size(); k++) {
          int SS = mk[j][k];
          f[cur][S & SS] = min(f[cur][S & SS], f[cur][S] + a[j]);
        }
    }
  }
  printf("%d", f[n & 1][0]);
}
