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.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.security.KeyStore.Builder;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;


public class LuckyPermutation {
	static StringTokenizer st = null;
    static BufferedReader reader = null;
    static int perm[] = null;
    static boolean used[] = null;
    static long fact[] = null;
    static void build(int pos, long k) throws Exception {
    	if (pos == perm.length)
    		return;
    	for (int i = 0; i < perm.length; ++i) {
    		if (!used[i]) {
    			if (k >= fact[perm.length - 1- pos]) {
    				k -= fact[perm.length - 1- pos];
    			} else {
    				perm[pos] = i;
    				used[i] = true;
    				build(pos + 1, k);
    				return;
    			}
    		}
    	}
    	throw new Exception();
    }
	public static void main(String[] args) throws Exception {	
		reader = new BufferedReader(new InputStreamReader(System.in));
		//reader = new BufferedReader(new FileReader("input.txt"));
		int n = NextInt();
		long k = (long)NextInt();
		int length = Math.min(13, n);
		fact = new long [20];
		fact[0] = 1;
		for (int i = 1; i < 20; ++i)
			fact[i] = (long)(i) * fact[i - 1];
		k -= 1;
		if (k >= fact[length]) {
			System.out.println("-1");
			return;
		}
		perm = new int[length];
		used = new boolean[length];
		System.err.println(Arrays.toString(fact));
		build(0, k);
		System.err.println(Arrays.toString(perm));
		long a[] = new long [3000];
		int p = 0;
		for (int len = 1; len <= 10; ++len) {
			for (int mask = 0; mask < (1 << len); ++mask) {
				long num = 0;
				for (int i = 0; i < len; ++i)
					num = num * 10 + 4 + 3 * ((mask & (1 << i)) == 0 ? 1 : 0);
				a[p++] = num;
			}
		}
		a = Arrays.copyOf(a, p);
		Arrays.sort(a);
		int res = 0;
		boolean good[] = new boolean[length];
		for (int i = 0; i < p; ++i) {
			if (a[i] <= n - length)
				res += 1;
			else
				break;
		}
		for (int i = 0; i < p; ++i) {
			if (a[i] >= n - length + 1 && a[i] <= n) {
				good[(int) (a[i] - n + length - 1)] = true;
			}
		}
		for (int i = 0; i < length; ++i) {
			if (good[i] && good[perm[i]])
				++res;
		}
		System.out.println(res);
	}
	static String NextToken() throws IOException {
		while(st == null || !st.hasMoreTokens())
			st = new StringTokenizer(reader.readLine());
		return st.nextToken();
	}
	static int NextInt() throws NumberFormatException, IOException {
		return Integer.parseInt(NextToken());
	}
}