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;
vector<long long> fact;
vector<long long> lucky;
void gen(int n, long long m) {
  lucky.push_back(m);
  if (n > 0) {
    gen(n - 1, m * 10 + 4);
    gen(n - 1, m * 10 + 7);
  }
}
void init() {
  lucky.clear();
  gen(10, 0);
  sort(lucky.begin(), lucky.end());
  fact.push_back(1);
  for (int i = 1; i < 20; ++i) {
    fact.push_back(i * fact.back());
  }
}
vector<long long> gao(vector<long long> v, long long m) {
  if (m == 0) {
    return v;
  } else {
    int n = (int)v.size() - 1;
    int k = m / fact[n];
    vector<long long> ret(1, v[k]);
    v.erase(v.begin() + k);
    v = gao(v, m % fact[n]);
    ret.insert(ret.end(), v.begin(), v.end());
    return ret;
  }
}
int main() {
  vector<long long> v, w;
  long long n, m;
  int ans;
  init();
  cin >> n >> m;
  ans = 0;
  if (n < 18) {
    if (m > fact[n]) {
      puts("-1");
      return 0;
    }
  } else {
    ans += upper_bound(lucky.begin(), lucky.end(), n - 18) - lucky.begin() - 1;
  }
  for (int i = 0; i < 18 && i < n; ++i) {
    v.push_back(n - i);
    w.push_back(n - i);
  }
  reverse(v.begin(), v.end());
  reverse(w.begin(), w.end());
  v = gao(v, m - 1);
  for (int i = 0; i < 18 && i < n; ++i) {
    if (binary_search(lucky.begin(), lucky.end(), v[i]) &&
        binary_search(lucky.begin(), lucky.end(), w[i])) {
      ++ans;
    }
  }
  printf("%d\n", ans);
  return 0;
}
