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 maxn = 1e3 + 10;
const int inf = 1e8;
int a[5], mp[maxn], dp[maxn][1 << 12], ans = inf;
char s[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= 4; i++) {
    scanf("%s", s + 1);
    for (int j = 1; j <= n; j++) mp[j] = (mp[j] << 1) + (s[j] == '.' ? 1 : 0);
  }
  for (int i = 0; i <= n + 1; i++)
    for (int j = 0; j < (1 << 12); j++) dp[i][j] = inf;
  int s = ((mp[1] << 4) + mp[2] << 4) + mp[3];
  dp[1][s] = 0;
  for (int i = 1; i <= n; i++) {
    for (int now = 0, nxt; now < (1 << 12); now++) {
      nxt = now | (1 << 8);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[1]);
      nxt = now | (1 << 9);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[1]);
      nxt = now | (1 << 10);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[1]);
      nxt = now | (1 << 11);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[1]);
      nxt = now | (1 << 8) | (1 << 9) | (1 << 4) | (1 << 5);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[2]);
      nxt = now | (1 << 9) | (1 << 10) | (1 << 5) | (1 << 6);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[2]);
      nxt = now | (1 << 10) | (1 << 11) | (1 << 6) | (1 << 7);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[2]);
      nxt = now | (1 << 8) | (1 << 9) | (1 << 10) | (1 << 4) | (1 << 5) |
            (1 << 6) | (1 << 0) | (1 << 1) | (1 << 2);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[3]);
      nxt = now | (1 << 9) | (1 << 10) | (1 << 11) | (1 << 5) | (1 << 6) |
            (1 << 7) | (1 << 1) | (1 << 2) | (1 << 3);
      dp[i][nxt] = min(dp[i][nxt], dp[i][now] + a[3]);
    }
    for (int now = 0, nxt; now < (1 << 12); now++) {
      nxt = ((now & ((1 << 8) - 1)) << 4) | mp[i + 3];
      if (now >> 8 == (1 << 4) - 1)
        dp[i + 1][nxt] = min(dp[i + 1][nxt], dp[i][now]);
      dp[i + 1][(1 << 12) - 1] =
          min(dp[i + 1][(1 << 12) - 1], dp[i][now] + a[4]);
    }
  }
  for (int now = 0; now < (1 << 12); now++) ans = min(ans, dp[n + 1][now]);
  printf("%d\n", ans);
  return 0;
}
