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;
set<long long> s;
void gen(long long i) {
  if (i > 1000000000ll) return;
  long long i1 = i * 10 + 4;
  long long i2 = i * 10 + 7;
  if (i1 <= 1000000000ll) s.insert(i1);
  gen(i1);
  if (i2 <= 1000000000ll) s.insert(i2);
  gen(i2);
}
long long fact[100];
int buff[100];
void simu(int n, int k) {
  for (int i = 0; i < n; i++) {
    int rest = n - i - 1;
    if (fact[rest] >= k) continue;
    int j = 0;
    while (k - fact[rest] > 0) {
      k -= fact[rest];
      j++;
    }
    swap(buff[i], buff[i + j]);
    sort(&buff[i + 1], &buff[n]);
  }
}
int main() {
  gen(0);
  vector<int> lucky(s.begin(), s.end());
  int ans = 0;
  int n, k;
  cin >> n >> k;
  fact[0] = 1;
  for (int i = 0; i < 100 - 1; i++) {
    if (fact[i] < 1000000000ll) {
      fact[i + 1] = (i + 1) * fact[i];
    } else {
      fact[i + 1] = fact[i];
    }
  }
  if (n < 100 && fact[n] < k) {
    puts("-1");
    return 0;
  }
  int b0;
  int nn;
  if (n < 100) {
    for (int i = 0; i < n; i++) {
      buff[i] = i + 1;
    }
    nn = n;
    b0 = buff[0];
    simu(n, k);
  } else {
    for (int i = 0; i < 100; i++) {
      buff[i] = n - (99 - i);
    }
    nn = 100;
    b0 = buff[0];
    simu(100, k);
  }
  for (int i = 0; i < (int)lucky.size(); i++) {
    int l = lucky[i];
    if (l > n) break;
    if (l >= b0) {
      for (int i = 0; i < nn; i++) {
        if (buff[i] == l) {
          int pos = b0 + i;
          ans += s.count(pos);
        }
      }
    } else {
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
