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 R = 4, C = 1024, B = 1 << 12, inf = 1e9 + 7;
char s[R][C];
int dp[C][B];
int main() {
  int c;
  scanf("%d", &c);
  int cost[4];
  for (int i = 0; i < 4; ++i) scanf("%d", cost + i);
  for (int i = 0; i < 4; ++i) {
    scanf("%s", s[i]);
    for (int j = 0; j < 5; j++) s[i][c + j] = '.';
  }
  for (int i = 0; i < C; ++i)
    for (int j = 0; j < B; ++j) dp[i][j] = inf;
  vector<pair<int, int> > tra;
  for (int d = 1; d <= 4; ++d) {
    for (int f = 0; f + d <= 4; ++f) {
      int mask = 0;
      for (int i = 0; i < d; ++i) {
        for (int j = 0; j < d; ++j) {
          mask |= 1 << (j * 4 + i + f);
        }
        tra.emplace_back((B - 1) ^ mask, cost[d - 1]);
      }
    }
  }
  int awal = 0;
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < 3; j++) {
      if (s[i][j] == '.') continue;
      awal |= 1 << (j * 4 + i);
    }
  }
  dp[0][awal] = 0;
  for (int i = 0; i <= c; ++i) {
    for (int mask = B - 1; mask >= 0; --mask) {
      if (dp[i][mask] >= inf) continue;
      dp[i + 1][0] = min(dp[i + 1][0], dp[i][mask] + cost[3]);
      for (auto it : tra) {
        int nmask = mask & it.first;
        if (nmask == mask) continue;
        dp[i][nmask] = min(dp[i][nmask], dp[i][mask] + it.second);
      }
      if (mask & 15) continue;
      int nmask = mask >> 4;
      for (int j = 0; j < 4; j++) {
        if (s[j][i + 3] == '.') continue;
        nmask |= 1 << (8 + j);
      }
      dp[i + 1][nmask] = min(dp[i + 1][nmask], dp[i][mask]);
    }
  }
  printf("%d\n", dp[c][0]);
  return 0;
}
