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;
int dp[5][65540], N, a[5], s[1105], mt[5][5];
char ss[5][1105];
void preWork() {
  for (int i = 1; i <= N; i++) {
    if (ss[1][i] == '.') s[i] += 1;
    if (ss[2][i] == '.') s[i] += 2;
    if (ss[3][i] == '.') s[i] += 4;
    if (ss[4][i] == '.') s[i] += 8;
  }
  s[N + 1] = s[N + 2] = s[N + 3] = s[N + 4] = 15;
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= 4 - i + 1; j++) {
      int bas = 0;
      for (int k = 1; k <= j; k++) bas += 1 << (i + k - 2);
      for (int k = 1; k <= j; k++) mt[i][j] += bas * (1 << ((k - 1) * 4));
    }
  }
}
void smin(int &x, int y) {
  if (x > y) x = y;
}
void solve() {
  int now = 4, nxt = (now + 1) % 5;
  memset(dp, 0x3f, sizeof(dp));
  dp[nxt][s[1] + s[2] * 16 + s[3] * 256 + s[4] * 4096] = 0;
  register int i, j, k, o;
  for (i = 1; i <= N; i++) {
    memset(dp[now], 0x3f, sizeof(dp[now]));
    now = nxt, nxt = (now + 1) % 5;
    int *f = dp[now];
    for (j = 0; j < 65536; j++) {
      if (f[j] == 0x3f3f3f3f) continue;
      if ((j & 15) == 15) {
        smin(dp[nxt][(j >> 4) + (s[i + 4] << 12)], f[j]);
        continue;
      }
      for (k = 1; k <= 4; k++) {
        for (o = 5 - k; o; o--) smin(f[j | mt[k][o]], f[j] + a[o]);
      }
    }
  }
  printf("%d", dp[nxt][65535]);
}
int main() {
  int aaaa = clock();
  scanf("%d", &N);
  for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= 4; i++) scanf("%s", ss[i] + 1);
  preWork();
  solve();
}
