Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

One day Petya dreamt of a lexicographically k-th permutation of integers from 1 to n. Determine how many lucky numbers in the permutation are located on the positions whose indexes are also lucky numbers.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 109) — the number of elements in the permutation and the lexicographical number of the permutation.

Output

If the k-th permutation of numbers from 1 to n does not exist, print the single number "-1" (without the quotes). Otherwise, print the answer to the problem: the number of such indexes i, that i and ai are both lucky numbers.

Examples

Input

7 4


Output

1


Input

4 7


Output

1

Note

A permutation is an ordered set of n elements, where each integer from 1 to n occurs exactly once. The element of permutation in position with index i is denoted as ai (1 ≤ i ≤ n). Permutation a is lexicographically smaller that permutation b if there is such a i (1 ≤ i ≤ n), that ai < bi, and for any j (1 ≤ j < i) aj = bj. Let's make a list of all possible permutations of n elements and sort it in the order of lexicographical increasing. Then the lexicographically k-th permutation is the k-th element of this list of permutations.

In the first sample the permutation looks like that:

1 2 3 4 6 7 5

The only suitable position is 4.

In the second sample the permutation looks like that:

2 1 3 4

The only suitable position is 4.
import java.util.*;
public class d {
public static void main(String[] args)
{
	Scanner input = new Scanner(System.in);
	int n = input.nextInt(), k = input.nextInt();
	if(n < 15)
	{
		long f = 1;
		for(int i = 2; i<=n; i++) f *= i;
		if(f < k)
		{
			System.out.println(-1);
			return;
		}
	}
	findLucky();
	int rest = n - 15;
	int res = countLuckyBelow(rest);
	long[] fs = new long[16];
	fs[0] = 1;
	for(int i = 1; i<fs.length; i++) fs[i] = fs[i-1] * i;
	int m = Math.min(n, 15);
	boolean[] luck = new boolean[m];
	for(int i = 0; i<m; i++)
	{
		if(lucky(n-i)) luck[m-1-i] = true;
	}
	boolean[] used = new boolean[m];
	for(int i = 0; i<m; i++)
	{
		long fact = fs[m-i-1];
		for(int j = 0; j<m; j++)
		{
			if(used[j]) continue;
			if(k > fact) k -= fact;
			else 
			{
				used[j] = true;
				if(luck[i] && lucky(n-m+1+j)) res++;
				break;
			}
		}
	}
	System.out.println(res);
}
static boolean lucky(int n)
{
	while(n>0)
	{
		if(n%10 != 4 && n%10 != 7) return false;
		n /= 10;
	}
	return true;
}
static void findLucky()
{
	lucky = new ArrayList<Integer>();
	lucky.add(4);
	lucky.add(7);
	ArrayList<Integer> cur = new ArrayList<Integer>();
	cur.add(4);
	cur.add(7);
	for(int i = 0; i<8; i++)
	{
		ArrayList<Integer> next = new ArrayList<Integer>();
		for(int x : cur)
		{
			next.add(10*x+4);
			next.add(10*x+7);
		}
		for(int x : next)
		{
			lucky.add(x);
		}
		cur = next;
	}
}
static ArrayList<Integer> lucky;
static int countLuckyBelow(int x)
{
	int res = 0;
	for(int k : lucky)
	{
		if(k <= x) res++; 
	}
	return res;
}
}
