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>
const int ROWS = 4;
const int MAX_N = 1005;
const int INF = 1e9;
using namespace std;
int n;
int cost[ROWS];
char grid[MAX_N][ROWS];
int orig_mask[MAX_N];
map<int, int> memo[MAX_N];
vector<int> first_masks[ROWS];
vector<int> clear_masks[ROWS];
int LOW_FOUR = (1 << 4) - 1;
int HIGH_FOUR = LOW_FOUR << 12;
int SIXTEEN = (1 << 16) - 1;
int dfs(int i, int mask) {
  if (i == 0 && mask == 0) {
    return 0;
  }
  if (memo[i].find(mask) == memo[i].end()) {
    int best = INF;
    if (i > 0 && (mask & HIGH_FOUR) == 0) {
      int shifted_mask = mask << 4;
      shifted_mask |= orig_mask[i - 1] & LOW_FOUR;
      best = min(best, dfs(i - 1, shifted_mask));
    } else {
      for (int sz = 0; sz < 4; ++sz) {
        for (int clear_mask : i == 0 ? first_masks[sz] : clear_masks[sz]) {
          if ((mask & clear_mask) == mask) continue;
          best = min(best, cost[sz] + dfs(i, mask & clear_mask));
        }
      }
    }
    memo[i][mask] = best;
  }
  return memo[i][mask];
}
int main() {
  scanf(" %d", &n);
  for (int i = 0; i < 4; ++i) {
    scanf(" %d", &cost[i]);
  }
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < n; ++j) {
      scanf(" %c", &grid[j][i]);
    }
  }
  for (int i = 0; i <= n - 4; ++i) {
    int mask = 0;
    for (int col = i; col < i + 4; ++col) {
      for (int row = 0; row < 4; ++row) {
        mask += (1 << (4 * (col - i) + row)) * (grid[col][row] == '*' ? 1 : 0);
      }
    }
    orig_mask[i] = mask;
  }
  for (int sz = 0; sz < 4; ++sz) {
    for (int i = 0; i < 4 - sz; ++i) {
      for (int j = 0; j < 4 - sz; ++j) {
        int mask = SIXTEEN;
        for (int col = i; col <= i + sz; ++col) {
          for (int row = j; row <= j + sz; ++row) {
            mask -= (1 << (4 * col + row));
          }
        }
        first_masks[sz].push_back(mask);
        if (i == 3 - sz) clear_masks[sz].push_back(mask);
      }
    }
  }
  printf("%d\n", dfs(n - 4, orig_mask[n - 4]));
  return 0;
}
