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;
bool islucky(int x) {
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  vector<long long> u;
  for (int(i) = 0; (i) < (int)(11); ++(i))
    if (i)
      for (int(mask) = 0; (mask) < (int)(1 << i); ++(mask)) {
        long long p = 0;
        for (int(j) = 0; (j) < (int)(i); ++(j)) {
          if (mask >> j & 1)
            p = p * 10 + 7;
          else
            p = p * 10 + 4;
        }
        u.push_back(p);
      }
  int len = 0;
  long long s = 1;
  while (s < K) s *= ++len;
  if (N < len) return puts("-1");
  vector<int> v;
  for (int(i) = 0; (i) < (int)(len); ++(i)) v.push_back(N - len + 1 + i);
  int ans = 0;
  for (int(i) = 0; (i) < (int)(len); ++(i)) {
    s /= len - i;
    int x = (K - 1) / s;
    if (islucky(N - len + 1 + i) && islucky(v[x])) ans++;
    v.erase(v.begin() + x);
    K -= x * s;
  }
  for (int(i) = 0; (i) < (int)(u.size()); ++(i))
    if (u[i] < (N - len)) ans++;
  printf("%d\n", ans);
}
