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;
long long f[30];
int gt(int l, long long n) {
  if (n >= l) return 0;
  int an = 0;
  an += gt(l, n * 10 + 4);
  an += gt(l, n * 10 + 7);
  return an + 1;
}
bool lk(int x) {
  bool ok = true;
  while (x) {
    int t = x % 10;
    if (t != 4 && t != 7) ok = false;
    x /= 10;
  }
  return ok;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  f[0] = 1;
  int i = 1;
  while (true) {
    f[i] = f[i - 1] * i;
    if (f[i] >= k) break;
    ++i;
  }
  if (i > n) {
    printf("-1");
    return 0;
  }
  int s = n - i + 1;
  int an = gt(s, 4) + gt(s, 7);
  vector<int> p;
  for (int i = s; i < n + 1; ++i) p.push_back(i);
  for (int i = s; i < n + 1; ++i) {
    int j = n - i;
    int z, t;
    if (f[j] <= k) {
      z = (k - 1) / f[j];
      t = p[z];
      k = (k - 1) % f[j] + 1;
    } else
      z = 0, t = p[0];
    p.erase(p.begin() + z);
    if (lk(t) && lk(i)) ++an;
  }
  printf("%d", an);
}
