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;
int c(long long a, long long b) {
  if (a > b) return 0;
  return 1 + c(a * 10 + 4, b) + c(a * 10 + 7, b);
}
int is(long long a) {
  return a == 0 || (a % 10 == 4 || a % 10 == 7) && is(a / 10);
}
long long f[20];
bool u[20];
int main() {
  int n, k;
  cin >> n >> k;
  k--;
  f[0] = 1;
  for (int i = 1; i <= 13; i++) f[i] = f[i - 1] * i;
  if (n <= 13 && f[n] <= k) {
    cout << -1;
    return 0;
  }
  int ans = c(0, n) - 1;
  for (int i = 13; i >= 0; i--) {
    int shift = k / f[i], c = -1;
    while (shift >= 0) {
      shift -= !u[++c];
    }
    u[c] = 1;
    int t = c - 13 + n;
    if (t > 0) {
      if (is(n - i)) ans += is(t) - 1;
    }
    k %= f[i];
  }
  cout << ans;
}
