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>
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
inline int read(T& x) {
  x = 0;
  char c = getchar(), f = 1;
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    if (c == -1) return -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  c *= f;
  return 1;
}
template <typename T, typename... Args>
inline int read(T& x, Args&... args) {
  return read(x) + read(args...);
}
template <typename T>
inline int write(T x) {
  if (x < 0) return putchar('-') & write((~x) + 1);
  if (x / 10) return write(x / 10) & putchar(x % 10 | 48);
  return putchar(x | 48);
}
template <typename T>
inline int write(T x, char c) {
  return write(x) + putchar(c);
}
inline int gotchar() {
  char c = getchar();
  while (c == ' ' || c == '\n') c = getchar();
  return c;
}
int n;
int cost[5];
int num[1005];
int dp[1005][(1 << 16) + 5];
bool m[10][10];
int sit[10][10];
signed main() {
  read(n);
  for (int i = 1; i <= 4; i++) read(cost[i]);
  for (int i = 0; i < 4; i++) {
    ;
    for (int j = 0; j < n; j++) {
      char c = gotchar();
      num[j] <<= 1, num[j] += (c == '*');
    }
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 1; j <= 4 - i; j++) {
      for (int k = 0; k < 4; k++)
        for (int x = 0; x < 4; x++) m[k][x] = 1;
      for (int k = 0; k < j; k++)
        for (int x = 0; x < j; x++) m[k][x + i] = 0;
      int b = 0;
      for (int k = 3; k >= 0; k--)
        for (int x = 0; x < 4; x++) b <<= 1, b |= m[k][x];
      sit[i][j] = b;
    }
  }
  memset(dp, 0x3f, sizeof dp);
  int inf = dp[0][0];
  int b = 0;
  for (int i = 3; i >= 0; i--) b <<= 4, b += num[i];
  dp[0][b] = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = (1 << 16) - 1; j >= 0; j--) {
      if (dp[i][j] == inf) continue;
      if (!(j & 0b1111)) {
        int s = (num[i + 4] << 12) | (j >> 4);
        dp[i + 1][s] = min(dp[i + 1][s], dp[i][j]);
      }
      for (int k = 0; k < 4; k++) {
        for (int x = 1; x <= 4 - k; x++) {
          dp[i][j & sit[k][x]] = min(dp[i][j & sit[k][x]], dp[i][j] + cost[x]);
        }
      }
    }
  }
  write(dp[n][0], '\n');
}
