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 good[3000];
long long fact[20];
int q = 0;
void generate(long long sum, int left) {
  if (!left) {
    good[q++] = sum;
  } else {
    if (!sum) {
      generate(sum, left - 1);
    }
    generate(sum * 10 + 4, left - 1);
    generate(sum * 10 + 7, left - 1);
  }
}
bool is_lucky(int v) {
  while (v) {
    if (v % 10 != 4 && v % 10 != 7) {
      return false;
    }
    v /= 10;
  }
  return true;
}
int per[20];
bool fre[20] = {};
int main() {
  generate(0, 10);
  int n;
  long long k;
  fact[0] = 1;
  for (int i = 1; i < 20; i++) {
    fact[i] = fact[i - 1] * i;
  }
  cin >> n >> k;
  if (n < 15) {
    if (k > fact[n]) {
      puts("-1");
      return 0;
    }
  }
  int ret = 0;
  for (int i = 1; i < q; i++) {
    if (good[i] <= n - 15) {
      ret++;
    }
  }
  int m = min(n, 15);
  k--;
  for (int i = 0; i < m; i++) {
    long long q = k / fact[m - i - 1];
    k -= q * fact[m - i - 1];
    for (int j = 0;; j++) {
      if (!fre[j]) {
        if (!q) {
          fre[j] = true;
          per[i] = j;
          break;
        }
        q--;
      }
    }
  }
  for (int i = 0; i < m; i++) {
    if (is_lucky(n - m + per[i] + 1) && is_lucky(n - m + i + 1)) {
      ret++;
    }
  }
  cout << ret << endl;
  return 0;
}
