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 + 5;
const int inf = 1e8;
int n;
int cost[10], a[10][maxn];
int p[10][10], apply[10];
int mask[66000][10], rep[maxn];
int mem[maxn][18][18][18];
map<int, int> val;
void gen(int row, int cur_val) {
  if (row == 4) {
    memset(p, 0, sizeof(p));
    for (int x = 0; x < 4; x++)
      for (int i = x; i < x + apply[x]; i++)
        for (int j = 0; j < apply[x]; j++) p[i][j] = 1;
    int id = 0;
    for (int j = 0; j < 4; j++)
      for (int i = 0; i < 4; i++) id = id * 2 + p[i][j];
    if (val.find(id) == val.end())
      val[id] = cur_val;
    else
      val[id] = min(val[id], cur_val);
    return;
  }
  for (int i = 0; i <= 4 - row; i++) {
    apply[row] = i;
    gen(row + 1, cur_val + cost[i]);
  }
}
int f(int x, int r1, int r2, int r3) {
  if (x == n + 1) return 0;
  if (mem[x][r1][r2][r3] == -1) {
    int ans = inf;
    for (auto temp : val) {
      int id = temp.first;
      if ((r1 | mask[id][1]) == 15) {
        ans =
            min(ans, temp.second + f(x + 1, r2 | mask[id][2], r3 | mask[id][3],
                                     rep[x + 3] | mask[id][4]));
      }
    }
    mem[x][r1][r2][r3] = ans;
  }
  return mem[x][r1][r2][r3];
}
int main() {
  char c;
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &cost[i]);
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= n; j++)
      scanf(" %c", &c), a[i][j] = ((c == '*') ? 0 : 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 4; j++) rep[i] = rep[i] * 2 + a[j][i];
  gen(0, 0);
  for (auto temp : val) {
    int id = temp.first;
    for (int i = 4; i >= 1; i--) {
      mask[temp.first][i] = id % 16;
      id /= 16;
    }
  }
  memset(mem, -1, sizeof(mem));
  printf("%d", f(1, rep[1], rep[2], rep[3]));
}
