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.*;
import java.util.*;
import java.math.*;

//created at 7:48 PM 10/27/11 by Abrackadabra

public class C {
    int IOMode = 0; //0 - consoleIO, 1 - <taskName>.in/out, 2 - input.txt/output.txt, 3 - test case generator
    String taskName = "";

    int l, n;
    int[] a;
    int[] f;

    void solve() throws IOException {
        n = nextInt();
        int k = nextInt() - 1;
        l = 100;
        a = new int[l + 1];
        f = new int[l + 1];
        for (int i = l; i >= 0; i--) {
            a[i] = n - l + i;
        }
        long fac = 1;
        int s = -1;
        for (int i = l; i >= 0; i--) {
            fac *= l - i + 1;
            f[i] = (int)fac;
            if (fac > k) {
                s = i;
                if (a[s] <= 0) {
                    out.println(-1);
                    return;
                }
                break;
            }
        }
        rotate(s, k);
        TreeSet<Long> g = new TreeSet<Long>();
        g.add((long)4);
        g.add((long)7);
        for (int i = 2; i < 12; i++) {
            TreeSet<Long> g1 = new TreeSet<Long>();
            for (Long t : g) {
                g1.add(t);
                g1.add(t * 10 + 4);
                g1.add(t * 10 + 7);
            }
            g = g1;
        }
        int r = 0;
        for (Long t : g) {
            if (t > n) break;
            long p = l - n + t;
            if (p < 0 || a[(int)p] <= 0) {
                r++;
            } else {
                if (g.contains((long)a[(int)p]))
                    r++;
            }
        }
        out.println(r);
    }

    void rotate(int s, int k) {
        if (s > l || k <= 0) return;
        int p = 0;
        while (k >= f[s + 1]) {
            k -= f[s + 1];
            p++;
        }
        for (int i = s + 1; i < l + 1; i++)
            if (i - s == p) {
                int t = a[s];
                a[s] = a[i];
                a[i] = t;
                break;
            }
        rotate(s + 1, k);
    }

    public static void main(String[] args) throws IOException {
        if (args.length > 0 && args[0].equals("Abra")) debugMode = true;
        new C().run();
    }

    long startTime = System.nanoTime(), tempTime = startTime, finishTime = startTime;
    long startMem = Runtime.getRuntime().totalMemory(), finishMem = startMem;

    void run() throws IOException {
        init();
        if (debugMode) {
            con.println("Start");
            con.println("Console output:");
        }
        solve();
        finishTime = System.nanoTime();
        finishMem = Runtime.getRuntime().totalMemory();
        out.flush();
        if (debugMode) {
            int maxSymbols = 1000, c = 0;
            BufferedReader tbr = new BufferedReader(new FileReader("input.txt"));
            char[] a = new char[maxSymbols];
            tbr.read(a);
            if (a[0] != 0) {
                con.println();
                con.println("File input:");
                con.print(a);
            }
            boolean left = true;
            for (int i = 0; i < maxSymbols; i++) left = left && a[i] != 0;
            if (left) con.println("...");
            else con.println();
            tbr = new BufferedReader(new FileReader("output.txt"));
            a = new char[maxSymbols];
            tbr.read(a);
            if (a[0] != 0) {
                con.println();
                con.println("File output:");
                con.print(a);
            }
            left = true;
            for (int i = 0; i < maxSymbols; i++) left = left && a[i] != 0;
            if (left) con.println("...");
            else con.println();
            con.println("Time passed:  " + (finishTime - startTime) / 1000000000.0 + " sec");
            con.println("Memory used:  " + (finishMem - startMem) + " bytes");
            con.println("Total memory: " + Runtime.getRuntime().totalMemory() + " bytes");
        }
    }

    boolean tick(double x) {
        if (System.nanoTime() - tempTime > x) {
            tempTime = System.nanoTime();
            con.println("Tick at " + (tempTime - startTime) / 1000000000 + " sec");
            con.print("   ");
            return true;
        }
        return false;
    }

    void printTime() {
        con.println((System.nanoTime() - tempTime) + " nanos passed");
        tempTime = System.nanoTime();
    }

    static boolean debugMode = false;
    PrintStream con = System.out;

    void init() throws IOException {
        if (debugMode) {
            br = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter(new FileWriter("output.txt"));
        } else
            switch (IOMode) {
                case 0:
                    br = new BufferedReader(new InputStreamReader(System.in));
                    out = new PrintWriter(System.out);
                    break;
                case 1:
                    br = new BufferedReader(new FileReader(taskName + ".in"));
                    out = new PrintWriter(new FileWriter(taskName + ".out"));
                    break;
                case 2:
                    br = new BufferedReader(new FileReader("input.txt"));
                    out = new PrintWriter(new FileWriter("output.txt"));
                    break;
                case 3:
                    out = new PrintWriter(new FileWriter("input.txt"));
                    break;
            }
    }

    BufferedReader br;
    PrintWriter out;
    StringTokenizer in;

    boolean hasMoreTokens() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            String line = br.readLine();
            if (line == null) return false;
            in = new StringTokenizer(line);
        }
        return true;
    }

    String nextString() throws IOException {
        return hasMoreTokens() ? in.nextToken() : null;
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextString());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextString());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextString());
    }
}
