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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int N = 1000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const long double eps = 1e-9;
const long double pi = atan2(0, -1);
const int ABC = 26;
string a[4];
int dp[N + 1][1 << 12];
int dp1[1 << 16];
vector<int> cost[4];
void init() {
  cost[0] = {65534, 65533, 65531, 65527};
  cost[1] = {65484, 65433, 65331};
  cost[2] = {63624, 61713};
  cost[3] = {0};
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  init();
  int n;
  cin >> n;
  int a1, a2, a3, a4;
  cin >> a1 >> a2 >> a3 >> a4;
  for (int i = 0; i < 4; i++) cin >> a[i];
  for (int i = 0; i < n + 1; i++)
    for (int mask = 0; mask < (1 << 12); mask++) dp[i][mask] = inf;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int nm = 0; nm < (1 << 16); nm++) dp1[nm] = inf;
    for (int mask = 0; mask < (1 << 12); mask++) {
      if (dp[i][mask] == inf) continue;
      int nm = mask;
      for (int j = 3; j > -1; j--) {
        nm <<= 1;
        if (a[j][i] == '*') nm++;
      }
      dp1[nm] = min(dp1[nm], dp[i][mask]);
    }
    for (int mask = (1 << 16) - 1; mask > -1; mask--) {
      if (i + 1 >= 1) {
        for (int j = 0; j < 4; j++)
          dp1[mask & cost[0][j]] = min(dp1[mask & cost[0][j]], dp1[mask] + a1);
      }
      if (i + 1 >= 2) {
        for (int j = 0; j < 3; j++)
          dp1[mask & cost[1][j]] = min(dp1[mask & cost[1][j]], dp1[mask] + a2);
      }
      if (i + 1 >= 3) {
        for (int j = 0; j < 2; j++)
          dp1[mask & cost[2][j]] = min(dp1[mask & cost[2][j]], dp1[mask] + a3);
      }
      if (i + 1 >= 4) {
        for (int j = 0; j < 1; j++)
          dp1[mask & cost[3][j]] = min(dp1[mask & cost[3][j]], dp1[mask] + a4);
      }
    }
    for (int mask = 0; mask < (1 << 12); mask++) dp[i + 1][mask] = dp1[mask];
  }
  cout << dp[n][0] << "\n";
  return 0;
}
