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;
constexpr int rsz = 5;
constexpr int sz = 1005;
int dp[rsz][sz][rsz][rsz][rsz][rsz], n, cost[rsz];
string s[rsz];
int mxs(const int& r) { return max(0, r - 1); }
int solve(int r = 0, int c = 0, int r1 = 0, int r2 = 0, int r3 = 0,
          int r4 = 0) {
  if (c == n) return 0;
  if (r == 4) return solve(0, c + 1, mxs(r1), mxs(r2), mxs(r3), mxs(r4));
  int& ret = dp[r][c][r1][r2][r3][r4];
  if (ret != 0x3f3f3f3f) return ret;
  int cur_cover[] = {r1, r2, r3, r4};
  int pre_cover[] = {r1, r2, r3, r4};
  if (s[r][c] == '.' || cur_cover[r]) ret = solve(r + 1, c, r1, r2, r3, r4);
  for (int i = 1; i <= 4; ++i) {
    for (int k = r; k < min(r + i, 4); ++k) {
      cur_cover[k] = max(cur_cover[k], i);
    }
    ret = min(ret, cost[i] + solve(r + 1, c, cur_cover[0], cur_cover[1],
                                   cur_cover[2], cur_cover[3]));
    for (int k = r; k < min(r + i, 4); ++k) {
      cur_cover[k] = pre_cover[k];
    }
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 1; i <= 4; ++i) {
    cin >> cost[i];
  }
  for (int i = 0; i < 4; ++i) {
    cin >> s[i];
  }
  memset(dp, 0x3f, sizeof dp);
  cout << solve();
}
