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;
char s[5][1005];
int a[5], b[5];
int dp1[1005][4096];
vector<pair<int, int> > v;
int code[1005];
int mn[4096];
void dfs(int i, int state, int cost) {
  if (i == 4) {
    mn[state] = min(mn[state], cost);
    return;
  }
  int mx = 3;
  if (i >= 2) mx--;
  if (i >= 3) mx--;
  for (int j = 0; j <= mx; j++) dfs(i + 1, state | (b[j] << i), cost + a[j]);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
  for (int i = 0; i < 4; i++) scanf("%s", s[i] + 1);
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 4096; j++) dp1[i][j] = 1e9, mn[j] = 1e9;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 4; j++) {
      if (s[j][i] == '*') code[i] += 1 << j;
    }
  }
  b[0] = 0;
  b[1] = 1;
  b[2] = 51;
  b[3] = 1911;
  dfs(0, 0, 0);
  for (int i = 0; i < 4096; i++) {
    if (mn[i] != 1e9) v.push_back(pair<int, int>(i, mn[i]));
  }
  dp1[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    int val1 = 1e9, val2 = 1e9;
    for (int j = 0; j < 4096; j++) {
      if (i > 3) val1 = min(val1, dp1[i - 4][j]);
      val2 = min(val2, dp1[i - 1][j]);
      if (dp1[i - 1][j] == 1e9) continue;
      for (int k = 0; k < v.size(); k++) {
        int nt = (j >> 4) | v[k].first;
        bool ok = (nt | code[i]) == nt;
        if (ok) dp1[i][nt] = min(dp1[i][nt], dp1[i - 1][j] + v[k].second);
      }
    }
    if (val1 != 1e9) dp1[i][15] = min(dp1[i][15], val1 + a[4]);
    if (val2 != 1e9) dp1[i][4095] = min(dp1[i][4095], val2 + a[4]);
  }
  int ans = 1e9;
  for (int j = 0; j < 4096; j++) ans = min(ans, dp1[n][j]);
  printf("%d\n", ans);
}
