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;
int n;
int a[5];
int dp[1005][(1 << 12)];
string s[5];
int b[1005];
inline void update(int x, int y, int val) {
  if (dp[x][y] < val) return;
  dp[x][y] = val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= 4; ++i) cin >> a[i];
  for (int i = 1; i <= 4; ++i) cin >> s[i];
  for (int i = 1; i <= n; ++i) {
    if (s[1][i - 1] == '.') b[i] |= 1;
    if (s[2][i - 1] == '.') b[i] |= 2;
    if (s[3][i - 1] == '.') b[i] |= 4;
    if (s[4][i - 1] == '.') b[i] |= 8;
  }
  int op = b[1];
  op |= (b[2] << 4);
  op |= (b[3] << 8);
  for (int i = 1; i <= n + 1; ++i)
    for (int j = 0; j < (1 << 12); ++j) dp[i][j] = 1e9;
  dp[1][op] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (1 << 12); ++j)
      if (dp[i][j] < 1e9) {
        int msk = (j >> 4);
        if (i + 3 <= n) msk |= (b[i + 3] << 8);
        if ((j & 15) == 15) update(i + 1, msk, dp[i][j]);
        int msk2 = 0;
        if (i + 3 <= n) {
          msk2 = (1 << 12) - 1;
          update(i + 1, msk2, dp[i][j] + a[4]);
        }
        msk = j;
        if (i + 2 <= n) {
          msk2 = msk;
          msk2 |= 7;
          msk2 |= 16;
          msk2 |= 32;
          msk2 |= 64;
          msk2 |= (1 << 8);
          msk2 |= (1 << 9);
          msk2 |= (1 << 10);
          update(i, msk2, dp[i][j] + a[3]);
          msk2 = msk;
          msk2 |= 2;
          msk2 |= 4;
          msk2 |= 8;
          msk2 |= 32;
          msk2 |= 64;
          msk2 |= 128;
          msk2 |= (1 << 9);
          msk2 |= (1 << 10);
          msk2 |= (1 << 11);
          update(i, msk2, dp[i][j] + a[3]);
        }
        if (i + 1 <= n) {
          msk2 = msk;
          msk2 |= 1;
          msk2 |= 2;
          msk2 |= 16;
          msk2 |= 32;
          update(i, msk2, dp[i][j] + a[2]);
          msk2 = msk;
          msk2 |= 2;
          msk2 |= 4;
          msk2 |= 32;
          msk2 |= 64;
          update(i, msk2, dp[i][j] + a[2]);
          msk2 = msk;
          msk2 |= 4;
          msk2 |= 8;
          msk2 |= 64;
          msk2 |= 128;
          update(i, msk2, dp[i][j] + a[2]);
        }
        msk2 = msk;
        msk2 |= 1;
        update(i, msk2, dp[i][j] + a[1]);
        msk2 = (msk | 2);
        update(i, msk2, dp[i][j] + a[1]);
        msk2 = (msk | 4);
        update(i, msk2, dp[i][j] + a[1]);
        msk2 = (msk | 8);
        update(i, msk2, dp[i][j] + a[1]);
      }
  }
  cout << dp[n + 1][0];
}
