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.
#include <bits/stdc++.h>
using namespace std;
int n, m, N, x, i, s, b[22];
long long f[15], a[2222];
bool u[15];
set<long long> st;
void rec(int l, long long s) {
  if (l > 0) {
    a[n++] = s;
    st.insert(s);
  }
  if (l < 10) rec(l + 1, s * 10 + 4);
  if (l < 10) rec(l + 1, s * 10 + 7);
}
void build(int l, int x) {
  if (l == 0) return;
  for (int j = 0; j < x; j++)
    if (!u[j]) {
      if (f[l - 1] >= m) {
        u[j] = true;
        b[x - l + 1] = j + 1;
        build(l - 1, x);
        break;
      } else
        m -= f[l - 1];
    }
}
int main() {
  rec(0, 0);
  sort(a, a + n);
  N = n;
  scanf("%d%d", &n, &m);
  for (f[0] = i = 1; i < 15; i++) f[i] = f[i - 1] * i;
  if (n < 15 && m > f[n]) {
    puts("-1");
    return 0;
  }
  build(min(14, n), min(14, n));
  if (n <= 14) {
    if (b[4] == 4) s++;
    if (b[7] == 7) s++;
  } else {
    x = n - 14;
    for (i = 0; i < N; i++)
      if (a[i] <= x) s++;
    for (i = 1; i <= 14; i++)
      if (st.count(x + i) && st.count(x + b[i])) s++;
  }
  printf("%d\n", s);
  return 0;
}
