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 int nmax = 1e3 + 42;
int dp[nmax][2 << 12];
char inp[4][nmax];
int c[5];
int n;
int get_state(int ind) {
  return (inp[0][ind] == '*') + (inp[1][ind] == '*') * 2 +
         (inp[2][ind] == '*') * 4 + (inp[3][ind] == '*') * 8;
}
int state_3_1 = (1 << 12) - 1 - (1 << 3) - (1 << 7) - (1 << 11);
int state_3_2 = (1 << 12) - 1 - (1 << 0) - (1 << 4) - (1 << 8);
int state_2_1 = (1 << 0) + (1 << 1) + (1 << 4) + (1 << 5);
int state_2_2 = (1 << 4) + (1 << 5) + (1 << 8) + (1 << 9);
int state_2_3 = (1 << 1) + (1 << 2) + (1 << 5) + (1 << 6);
int state_2_4 = (1 << 5) + (1 << 6) + (1 << 9) + (1 << 10);
int state_2_5 = (1 << 2) + (1 << 3) + (1 << 6) + (1 << 7);
int state_2_6 = (1 << 6) + (1 << 7) + (1 << 10) + (1 << 11);
int rec(int pos, int state) {
  if (pos > n) return 0;
  if (dp[pos][state] != -1) return dp[pos][state];
  if (state % 16 == 0) {
    dp[pos][state] = rec(pos + 1, state / 16 + get_state(pos + 3) * 16 * 16);
    return dp[pos][state];
  }
  int ret = c[4] + rec(pos + 4, get_state(pos + 4) + get_state(pos + 5) * 16 +
                                    get_state(pos + 6) * 16 * 16);
  if ((state & state_3_1))
    ret = min(ret, c[3] + rec(pos, (state | state_3_1) ^ state_3_1));
  if ((state & state_3_2))
    ret = min(ret, c[3] + rec(pos, (state | state_3_2) ^ state_3_2));
  if ((state & state_2_1))
    ret = min(ret, c[2] + rec(pos, (state | state_2_1) ^ state_2_1));
  if ((state & state_2_2))
    ret = min(ret, c[2] + rec(pos, (state | state_2_2) ^ state_2_2));
  if ((state & state_2_3))
    ret = min(ret, c[2] + rec(pos, (state | state_2_3) ^ state_2_3));
  if ((state & state_2_4))
    ret = min(ret, c[2] + rec(pos, (state | state_2_4) ^ state_2_4));
  if ((state & state_2_5))
    ret = min(ret, c[2] + rec(pos, (state | state_2_5) ^ state_2_5));
  if ((state & state_2_6))
    ret = min(ret, c[2] + rec(pos, (state | state_2_6) ^ state_2_6));
  for (int i = 0; i < 12; i++)
    if ((state & (1 << i))) {
      ret = min(ret, c[1] + rec(pos, state ^ (1 << i)));
    }
  dp[pos][state] = ret;
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n;
  for (int i = 1; i <= 4; i++) cin >> c[i];
  for (int i = 0; i < 4; i++)
    for (int j = 1; j <= n; j++) cin >> inp[i][j];
  cout << rec(1, get_state(1) + get_state(2) * 16 + get_state(3) * 16 * 16)
       << endl;
  return 0;
}
