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;
namespace TYC {
const int N = 1010, MX = (1 << 16) - 1, inf = 0x3f3f3f3f;
int n, cost[6], row[N], bin[20], num[6][6], dp[N][MX + 100];
vector<int> mark[6];
char str[6][N];
inline int square(int siz, int r) {
  int ans = 0;
  for (int i = 0; i < siz; i++)
    for (int j = 1; j <= siz; j++) ans |= bin[num[r + i][j]];
  return ans;
}
void init() {
  bin[1] = 1;
  for (int i = 2; i <= 16; i++) bin[i] = bin[i - 1] << 1;
  int tot = 0;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= 4; j++) num[j][i] = ++tot;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 4; j++)
      if (str[j][i] == '*') row[i] |= bin[j];
  for (int siz = 1; siz <= 4; siz++)
    for (int i = 1; i + siz - 1 <= 4; i++)
      mark[siz].push_back(MX ^ square(siz, i));
}
void work() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &cost[i]);
  for (int i = 1; i <= 4; i++) scanf("%s", str[i] + 1);
  init();
  int fir = 0, tmp = 0;
  for (int i = 1; i <= min(4, n); i++) fir = fir | (row[i] << ((i - 1) * 4));
  for (int i = 1; i <= 12; i++) tmp |= bin[i];
  memset(dp, inf, sizeof(dp));
  dp[1][fir] = 0;
  for (int i = 1; i <= n; i++)
    for (int s = MX; s >= 0; s--)
      if (dp[i][s] != inf) {
        if (!(s & 15)) {
          int d = (s >> 4) | (row[i + 4] << 12);
          dp[i + 1][d] = min(dp[i + 1][s], dp[i][s]);
        }
        for (int siz = 1; i + siz - 1 <= n; siz++)
          for (int j = 0; j < mark[siz].size(); j++) {
            int k = mark[siz][j];
            if ((s & k) != s)
              dp[i][s & k] = min(dp[i][s & k], dp[i][s] + cost[siz]);
          }
      }
  printf("%d\n", dp[n][0]);
}
}  // namespace TYC
int main() {
  TYC::work();
  return 0;
}
