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 LuckyPermutation {
	public static void main (String [] arg) {
		Scanner sc = new Scanner(System.in);
		long N = sc.nextInt();
		long K = sc.nextInt()-1;
		boolean overflow = (N > 15);
		int [] P = (overflow) ? permutation(15, K) : permutation((int)N,K);
		
		if (P == null) { System.out.println("-1"); return;}
		if (overflow)
			for (int i = 0; i<P.length; ++i) P[i] += (N-15);
			
		//System.err.println(Arrays.toString(P));
		
		long count = 0;
		for (int i = 0; i<P.length; ++i) {
			long index = (overflow) ? (N-15)+i+1 : i+1;
			//System.err.println("Trying " + index + " and " + P[i]);
			if (isLucky(index) && isLucky(P[i])) count++;
		}
		ArrayList<Long> Lucky = new ArrayList<Long>(1<<12);
		Lucky.add(4L);
		Lucky.add(7L);
		for (int LEN = 2; LEN <= 10; ++LEN) {
			for (int i = 0; i<(1<<LEN); ++i) {
				long NUM = 0;
				for (int j = 0; j<LEN; ++j) {
					NUM += ((i & (1L<<j)) != 0) ? 4 : 7;
					NUM *= 10L;
				}
				Lucky.add(NUM/10);
			}
		}
		Collections.sort(Lucky);
		
		int start = 0;
		long L = N-15;
		for (int i = 0; i<Lucky.size(); ++i) if (L < Lucky.get(i)) {start=i; break;}
		count += (start);
		System.out.println(count);
	}
	public static boolean isLucky(long i) {
		//System.err.println(i + " : " + Long.toString(i).replaceAll("[47]",""));
		return (Long.toString(i).replaceAll("[47]","").length() == 0);
	}
	
	
	public static int [] permutation (int N, long kk) {
		long tA = kk;
		long tF = fact(N);
		if (kk >= tF) return null;
		boolean [] used = new boolean [N];
		int [] B = new int [N];
		for (int i = B.length-1; i>0; --i) {
			tF /= (i+1);
			long k = tA / tF;
			int j = 0;
			for (;j<B.length; ++j) {
				if (used[j]) continue;
				if (--k < 0) break;
			}
			used[j] = true;
			B[B.length-1-i] = j+1;
			
			tA %= tF;
		}
		
		for (int j = 0; j<B.length; ++j) if (!used[j]) {B[B.length-1]=j+1;break;}	
		return B;
	}
	
	public static long fact (long N) {
		long F = 1;
		for (long i = 2; i<=N; ++i) F*=i;
		return F;
	}

}