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.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.util.Arrays;
import java.io.PrintWriter;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Collection;
import java.io.InputStream;
import java.util.Vector;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author bkand1908
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskE solver = new TaskE();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskE {
    public void solve(int testNumber, Scanner in, PrintWriter out) {

        int count = in.nextInt();
        int index = in.nextInt() - 1;
        int tail = -1;

        for (int i = 0; i <= count; ++i)
            if (factorial(i) > index) {
                tail = i;
                break;
            }
        if (tail == -1) {
            out.println(-1);
            return;
        }

        long[] lucky = generateLucky();

        int ans = Arrays.binarySearch(lucky, count - tail);
        if (ans < 0)
            ans = -ans - 1;
        else
            ++ans;

        boolean[] taken = new boolean[tail];
        int start = count - tail + 1;
        for (int i = 0; i < tail; ++i) {
            long fact = factorial(tail - i - 1);
            for (int j = 0; j < tail; ++j)
                if (!taken[j]) {
                    if (fact > index) {
                        taken[j] = true;
                        if (Arrays.binarySearch(lucky, start + i) >= 0 &&
                                Arrays.binarySearch(lucky, start + j) >= 0)
                            ++ans;
                        break;
                    } else
                        index -= fact;
                }
        }

        out.println(ans);
    }

    private long factorial(int n) {
        long res = 1;
        for (long i = 2; i <= n; ++i)
            res *= i;
        return res;
    }


    private long[] generateLucky() {
        Vector<Long> lucky = new Vector<Long>();
        Queue<Long> q = new LinkedList<Long>();

        q.add(4l);
        q.add(7l);
        while (!q.isEmpty()) {
            long value = q.remove();
            if (value >= 1000000000)
                continue;
            lucky.add(value);
            q.add(value * 10 + 4);
            q.add(value * 10 + 7);
        }

        long[] ans = new long[lucky.size()];
        for (int i = 0; i < ans.length; ++i)
            ans[i] = lucky.elementAt(i);
        return ans;
    }
}

