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 INF = 0x3f3f3f3f;
const int MX = 1e3 + 5;
const int MM = (1 << 16) - 1;
int n, a[5];
int dp[MX][MM + 5], r[MX];
char mp[4][MX];
vector<int> mark[5];
int get_sta(int l, int x) {
  int ret = 0;
  for (int i = 0; i < x; i++) {
    for (int j = 0; j < x; j++) {
      ret |= 1 << (4 * j + l + i);
    }
  }
  return ret;
}
int get_row(int i) {
  int ret = 0;
  for (int j = 0; j < 4; j++)
    if (mp[j][i] == '*') ret |= 1 << j;
  return ret;
}
void init() {
  for (int i = 1; i <= 4; i++) {
    for (int j = 0; j + i <= 4; j++) {
      mark[i].push_back(MM ^ get_sta(j, i));
    }
  }
  for (int i = 0; i < n; i++) r[i] = get_row(i);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
  for (int i = 0; i < 4; i++) scanf("%s", mp[i]);
  init();
  memset(dp, 0x3f, sizeof(dp));
  int S = 0;
  for (int i = 0; i < min(4, n); i++) S |= r[i] << (i * 4);
  dp[0][S] = 0;
  for (int i = 0; i < n; i++) {
    for (int S = MM; S >= 0; S--) {
      if (dp[i][S] == INF) continue;
      if ((S & 15) == 0) {
        int SS = S >> 4 | (r[i + 4] << 12);
        dp[i + 1][SS] = min(dp[i + 1][SS], dp[i][S]);
      }
      for (int j = 1; j <= 4 && i + j <= n; j++) {
        for (auto SS : mark[j]) {
          dp[i][S & SS] = min(dp[i][S & SS], dp[i][S] + a[j]);
        }
      }
    }
  }
  printf("%d\n", dp[n][0]);
  return 0;
}
