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];
int a[20];
long long fact() {
  f[0] = 1;
  for (int i = 1; i <= 15; i++) {
    f[i] = f[i - 1] * i;
  }
}
int sol(int n, int k) {
  int vt[20];
  for (int i = 1; i <= n; i++) {
    vt[i] = i;
  }
  int t = 1;
  int l;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (t + j * f[n - i] > k) {
        l = j - 1;
        break;
      }
    }
    t += l * f[n - i];
    a[i] = vt[l + 1];
    for (int j = l + 1; j <= n; j++) {
      vt[j] = vt[j + 1];
    }
  }
}
int check(int n) {
  if (n == 0) return 0;
  while (n != 0) {
    if (n % 10 != 4 && n % 10 != 7) return 0;
    n /= 10;
  }
  return 1;
}
int main() {
  fact();
  int n, k;
  cin >> n >> k;
  if (n < 14) {
    int res = 0;
    if (k > f[n])
      cout << -1;
    else {
      sol(n, k);
      for (int i = 1; i <= n; i++) {
        res += check(a[i]) * check(i);
      }
      cout << res;
    }
  } else {
    int res = 0;
    for (int i = 1; i <= n - 13; i++) {
      res += check(i);
    }
    sol(13, n);
    for (int i = 1; i <= 13; i++) {
      a[i] += n - 13;
    }
    for (int i = 1; i <= 13; i++) {
      res += check(a[i]) * check(n - 13 + i);
    }
    cout << res;
  }
}
