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 maxN = 1005;
const int inf = 1e9;
int dp[maxN][(1 << 12) + 5];
char a[4][maxN];
int N;
int pric[6];
int val[6];
pair<int, int> b[125];
int main() {
  scanf("%d", &N);
  val[0] = pric[0] = 0;
  for (int i = 1; i <= 4; ++i) {
    scanf("%d", &pric[i]);
    val[i] = 0;
    for (int j = 0; j < i; ++j)
      for (int k = 0; k < i; ++k) val[i] |= (1 << (j * 4 + k));
  }
  int p = 0;
  for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 4; ++j) {
      for (int k = 0; k < 3; ++k) {
        for (int l = 0; l < 2; ++l) {
          int t = val[i] | (val[j] << 1) | (val[k] << 2) | (val[l] << 3);
          b[++p] = make_pair(pric[i] + pric[j] + pric[k] + pric[l], t);
        }
      }
    }
  }
  for (int i = 0; i < 4; ++i) scanf("%s", a[i]);
  for (int i = 0; i <= N; ++i)
    for (int j = 0; j < (1 << 12); ++j) dp[i][j] = inf;
  int s = 0;
  for (int i = 0; i < 3; ++i)
    for (int j = 0; j < 4; ++j)
      if (a[j][i] == '.') s += (1 << (i * 4 + j));
  dp[0][s] = 0;
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < (1 << 12); ++j) {
      if (dp[i][j] == inf) continue;
      int new_ = 0;
      if (i + 3 < N) {
        for (int k = 0; k < 4; ++k)
          if (a[k][i + 3] == '.') new_ += 1 << (12 + k);
      }
      for (int k = 1; k <= 120; ++k) {
        int t2 = b[k].first, t1 = b[k].second;
        if (((j | t1) & ((1 << 4) - 1)) == ((1 << 4) - 1)) {
          int t = (j | new_ | t1) >> 4;
          dp[i + 1][t] = min(dp[i + 1][t], dp[i][j] + t2);
        }
      }
    }
  }
  printf("%d\n", dp[N][0]);
  return 0;
}
