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 N = 1005;
int n, a[4];
char s[4][N];
inline int ml(int i) {
  int ans = 0;
  for (int j = int(0); j < int(4); j++) ans |= (int(s[j][i]) << j);
  return ans;
}
vector<int> v;
inline int index(int x) {
  if (x == 1) return 0;
  if (x == 4) return 1;
  if (x == 9) return 2;
  if (x == 16) return 3;
  assert(0);
}
int dp[N][1 << 12];
int f(int i, int bit) {
  int &ans = dp[i][bit];
  if (~ans) return ans;
  ans = 1000000000;
  if (i == n && !bit) return ans = 0;
  int mbit = bit | (ml(i) << 12), nbit, id;
  if ((mbit & 15) == 0) ans = min(ans, f(i + 1, mbit >> 4));
  for (int j = int(0); j < int(v.size()); j++) {
    nbit = mbit & ~v[j];
    id = index(__builtin_popcount(v[j]));
    if (nbit == mbit) continue;
    if (nbit & 15)
      ans = min(ans, a[id] + f(i, nbit & ~0xf000));
    else
      ans = min(ans, a[id] + f(min(n, i + 1), nbit >> 4));
  }
  return ans;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%d", &n);
  for (int j = int(0); j < int(4); j++) scanf("%d", a + j);
  for (int j = int(0); j < int(4); j++) scanf(" %s", s[j]);
  for (int j = int(0); j < int(4); j++)
    for (int i = int(0); i < int(n); i++) s[j][i] = s[j][i] == '*';
  v.push_back(0x1), v.push_back(0x2), v.push_back(0x4), v.push_back(0x8);
  v.push_back(0x33), v.push_back(0x66), v.push_back(0xcc);
  v.push_back(0x777), v.push_back(0xeee);
  v.push_back(0xffff);
  printf("%d\n", f(3, ml(0) | (ml(1) << 4) | (ml(2) << 8)));
  return 0;
}
