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 INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 1e6 + 7;
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = 1ll * a * a % M)
    if (b & 1) ret = 1ll * ret * a % M;
  return ret;
}
int n, a1, a2, a3, a4;
int i, j, k;
int a[maxn];
int f[maxn], g[maxn];
int main() {
  scanf("%d%d%d%d%d", &n, &a1, &a2, &a3, &a4);
  for (k = 0; k < 4; k++)
    for (i = 0; i < n; i++) {
      char c;
      do c = getchar();
      while (c != '.' && c != '*');
      a[i] = a[i] * 2 + (c == '*');
    }
  for (i = 0; i < n; i++) {
    for (j = 0; j < (1 << 16); j++) g[j] = f[j], f[j] = INF;
    for (j = 0; j < (1 << 12); j++) f[j << 4 | a[i]] = g[j];
    for (j = (1 << 16) - 1; j >= 0; j--) {
      f[j & (~65535)] = min(f[j & (~65535)], f[j] + a4);
      f[j & (~1911)] = min(f[j & (~1911)], f[j] + a3);
      f[j & (~3822)] = min(f[j & (~3822)], f[j] + a3);
      f[j & (~51)] = min(f[j & (~51)], f[j] + a2);
      f[j & (~102)] = min(f[j & (~102)], f[j] + a2);
      f[j & (~204)] = min(f[j & (~204)], f[j] + a2);
      f[j & (~1)] = min(f[j & (~1)], f[j] + a1);
      f[j & (~2)] = min(f[j & (~2)], f[j] + a1);
      f[j & (~4)] = min(f[j & (~4)], f[j] + a1);
      f[j & (~8)] = min(f[j & (~8)], f[j] + a1);
    }
  }
  printf("%d\n", f[0]);
}
