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>
inline T read1() {
  T t = 0;
  char k = getchar();
  bool flag = 0;
  while ('0' > k || k > '9') {
    if (k == '-') flag = 1;
    k = getchar();
  }
  while ('0' <= k && k <= '9')
    t = (t << 3) + (t << 1) + (k ^ 48), k = getchar();
  return flag ? -t : t;
}
int s, x[5], dp[2][16][16][16];
char str[5][1005];
int main() {
  s = read1<long long>();
  for (int i = 1; i < 5; ++i) x[i] = read1<long long>();
  for (int i = 1; i <= 4; ++i) scanf("%s", str[i] + 1);
  memset(dp[0], 0x7f, sizeof(dp[0]));
  dp[0][15][15][15] = 0;
  for (int i = 1; i <= s; ++i) {
    memset(dp[i & 1], 0x7f, sizeof(dp[i & 1]));
    int t = 0;
    for (int j = 1; j < 5; ++j) t = t << 1 | (str[j][i] == '.');
    for (int j = 0; j < 16; ++j)
      for (int k = 0; k < 16; ++k)
        for (int l = 0; l < 16; ++l)
          dp[i & 1][15][15][15] =
              min(dp[i & 1][15][15][15], dp[~i & 1][j][k][l] + x[4]);
    for (int k = 0; k < 16; ++k)
      for (int l = 0; l < 16; ++l)
        dp[i & 1][k][l][t] = min(dp[i & 1][k][l][t], dp[~i & 1][15][k][l]);
    for (int j = 0; j < 16; ++j)
      for (int k = 0; k < 16; ++k)
        for (int r = 0; r < 16; ++r)
          for (int l = 1; l < 4; ++l)
            for (int w = 0; w < 5 - l; ++w) {
              int op[3] = {j, k, r};
              for (int f = 2; f > 2 - l; --f) op[f] |= (1 << l) - 1 << w;
              dp[i & 1][op[0]][op[1]][op[2]] = min(
                  dp[i & 1][op[0]][op[1]][op[2]], dp[i & 1][j][k][r] + x[l]);
            }
  }
  cout << dp[s & 1][15][15][15];
  return 0;
}
