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 MAX = 1e3 + 111;
const int INF = 1e9;
const int MSK = (1 << 16) + 1;
const int M = 2;
int n;
int a[4];
int dp[M][4][MSK], mm[MAX];
int tt[MSK][4][5];
string s[4];
void init() {
  for (int msk = 0; msk < (1 << 16); ++msk) {
    for (int fst = 0; fst < 4; ++fst) {
      for (int i = 0; i <= 4 - fst; ++i) {
        int nm = msk;
        for (int j = fst; j < fst + i; ++j) {
          for (int k = 0; k < i; ++k) {
            nm |= (1 << (j + 4 * k));
          }
        }
        tt[msk][fst][i] = nm;
      }
    }
  }
}
inline void go(int id, int msk, int pos) {
  for (int i = ((msk & (1 << pos)) ? 0 : 1); i <= min(n - id, 4 - pos); ++i) {
    int to = tt[msk][pos][i];
    int nid = id % M;
    int npos = pos + 1;
    if (pos == 3 || i == 4) {
      nid = (id + 1) % M;
      npos = 0;
      to >>= 4;
      to |= (mm[id + 4] << 12);
    }
    dp[id % M][pos][msk] = min(dp[nid][npos][to] + ((i == 0) ? 0 : a[i - 1]),
                               dp[id % M][pos][msk]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  init();
  cin >> n;
  for (int i = 0; i < 4; ++i) cin >> a[i];
  for (int i = 0; i < 4; ++i) {
    cin >> s[i];
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 3; j >= 0; --j) {
      mm[i] = mm[i] * 2 + (s[j][i] == '.');
    }
  }
  for (int i = 0; i < M; ++i) {
    for (int j = 0; j < 4; ++j) {
      fill(dp[i][j], dp[i][j] + (1 << 16), INF);
    }
  }
  dp[n % M][0][0] = 0;
  for (int id = n - 1; id >= 0; --id) {
    for (int beg = (1 << 12) - 1; beg >= 0; --beg) {
      for (int i = 3; i >= 0; --i) {
        int msk = beg + (mm[id + 3] << 12);
        dp[id % M][i][msk] = INF;
        go(id, msk, i);
      }
    }
  }
  int id = 0;
  int msk = mm[id] + mm[id + 1] * (1 << 4) + mm[id + 2] * (1 << 8) +
            mm[id + 3] * (1 << 12);
  cout << dp[id][0][msk];
  return 0;
}
