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> getkth(const vector<long long>& v, long long k) {
  long long n = v.size();
  if (n == 1) {
    assert(k == 1);
    return v;
  }
  long long fact = 1;
  for (int i = (int)(1); i <= (int)(n - 1); i++) fact *= i;
  long long pos = k / fact;
  if (k % fact) pos++;
  assert(pos >= 1);
  vector<long long> left;
  for (int i = 0; i < (int)(n); i++)
    if (i != (pos - 1)) left.push_back(v[i]);
  long long leftval = k % fact;
  if (leftval == 0) leftval = fact;
  vector<long long> ret = getkth(left, leftval);
  vector<long long> rett;
  rett.push_back(v[pos - 1]);
  for (int i = 0; i < (int)(ret.size()); i++) rett.push_back(ret[i]);
  return rett;
}
set<long long> S;
int anss = 0;
long long NUM;
void generate(long long i) {
  if (i > 1000000000) return;
  S.insert(i);
  if (i <= NUM - 13) anss++;
  generate(10 * i + 4);
  generate(10 * i + 7);
}
int main() {
  vector<long long> v;
  long long n, k;
  cin >> n >> k;
  NUM = n;
  generate(4);
  generate(7);
  long long f = 1;
  bool f1 = false;
  for (int i = (int)(1); i <= (int)(n); i++) {
    f *= i;
    if (f >= k) {
      f1 = true;
      break;
    }
  }
  if (!f1) {
    cout << -1 << endl;
    return 0;
  }
  if (n <= 13) {
    vector<long long> v;
    for (int i = (int)(1); i <= (int)(n); i++) v.push_back(i);
    vector<long long> ret = getkth(v, k);
    int ans = 0;
    for (int i = 0; i < (int)(n); i++) {
      if ((S.find(i + 1) != S.end()) && (S.find(ret[i]) != S.end())) ans++;
    }
    cout << ans << endl;
  } else {
    vector<long long> v;
    for (int i = (int)(1); i <= (int)(13); i++) v.push_back(i);
    vector<long long> ret = getkth(v, k);
    for (int i = 0; i < (int)(13); i++) ret[i] += (n - 13);
    int ans = 0;
    for (int i = 0; i < (int)(13); i++) {
      if ((S.find(n - 12 + i) != S.end()) && (S.find(ret[i]) != S.end())) ans++;
    }
    ans += anss;
    cout << ans << endl;
  }
  return 0;
}
