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;
vector<long long> v;
long long k, sum;
int n, a[15], res;
bool fr[15];
bool lucky(long long n) {
  while (n != 0) {
    if (n % 10 != 4 && n % 10 != 7) return false;
    n /= 10;
  }
  return true;
}
void gen(int k, int pos) {
  if (pos == k + 1) {
    v.push_back(sum);
    return;
  }
  sum = sum * 10 + 4;
  gen(k, pos + 1);
  sum /= 10;
  sum = sum * 10 + 7;
  gen(k, pos + 1);
  sum /= 10;
}
long long gt(int n) {
  long long res = 1;
  for (int i = (1), _b = (n); i <= _b; i++) res = res * i;
  return res;
}
int main() {
  sum = 0;
  for (int i = (1), _b = (10); i <= _b; i++) gen(i, 1);
  cin >> n >> k;
  int pos = max(0, n - 13);
  int i = 0;
  while (v[i] <= pos) i++;
  res = i;
  n = min(13, n);
  if (gt(n) < k) {
    cout << -1;
    return 0;
  }
  memset(fr, true, sizeof(fr));
  for (int i = (1), _b = (n); i <= _b; i++) {
    long long x = gt(n - i);
    for (int j = (1), _b = (n); j <= _b; j++)
      if (fr[j]) {
        if (x < k)
          k -= x;
        else {
          a[i] = j;
          fr[j] = false;
          break;
        }
      }
  }
  for (int i = (1), _b = (n); i <= _b; i++)
    if (lucky(pos + i) && lucky(pos + a[i])) res++;
  cout << res;
  return 0;
}
