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[20];
vector<int> L, E;
int N;
void get(long long num = 0, int digitos = 0) {
  if (num > N) return;
  if (digitos) L.push_back(num);
  get(num * 10 + 4, digitos + 1);
  get(num * 10 + 7, digitos + 1);
}
void Permutar(vector<int>& X, int k) {
  int tam = X.size();
  int quedan = tam - 1;
  for (int i = 0; i < tam; i++, quedan--) {
    sort(X.begin() + i, X.end());
    int avanzar = 0;
    while (k > F[quedan]) k -= F[quedan], avanzar++;
    swap(X[i], X[i + avanzar]);
  }
}
int main() {
  int P;
  F[0] = 1;
  for (int i = 0; i < 16; i++) F[i + 1] = (i + 1) * F[i];
  cin >> N >> P;
  get();
  sort(L.begin(), L.end());
  for (int i = N; E.size() < 15 && i > 0; i--) E.push_back(i);
  if (P > F[E.size()]) {
    cout << -1 << endl;
    return 0;
  }
  Permutar(E, P);
  int antes = N - E.size();
  int resp = 0;
  for (int i = 0; i < L.size(); i++) {
    if (L[i] <= antes) {
      resp++;
      continue;
    }
    int pos = find(E.begin(), E.end(), L[i]) - E.begin() + antes + 1;
    resp += binary_search(L.begin(), L.end(), pos);
  }
  cout << resp << endl;
  return 0;
}
