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 fact[40];
long long k, n, p, r, d;
long long t[20];
long long ans;
long long u;
bool luckychecker(long long A) {
  while (A > 0) {
    if (A % 10 != 7 && A % 10 != 4) {
      return 0;
    }
    A /= 10;
  }
  return 1;
}
void checker(long long A, long long B) {
  if (B == int(log10(n)) + 1) {
    if (A <= n) {
      ans++;
    }
  } else {
    checker(A * 10 + 4, B + 1);
    checker(A * 10 + 7, B + 1);
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i < 20; i++) {
    fact[i] = fact[i - 1] * i;
  }
  cin >> n >> k;
  if (k > fact[min(n, long long(18))]) {
    cout << k << ' ' << fact[n] << endl;
    cout << -1;
    return 0;
  }
  for (int i = 1; i < 15; i++) {
    if (fact[i] > k) {
      p = i;
      break;
    }
    if (fact[i] == k) {
      p = i;
      d = 1;
      break;
    }
  }
  for (long long i = n; i > n - p; i--) {
    t[n - i] = i;
  }
  sort(t, t + p);
  if (d == 1) {
    reverse(t, t + p);
    for (int i = 0; i < p; i++) {
      if (luckychecker(t[i]) + luckychecker(n - p + 1 + i) == 2) {
        ans++;
      }
    }
  } else {
    k--;
    for (int i = 0; i < p; i++) {
      r = k / fact[p - i - 1];
      k %= fact[p - i - 1];
      swap(t[i], t[i + r]);
      sort(t + i + 1, t + p);
      if (luckychecker(t[i]) + luckychecker(n - p + 1 + i) == 2) {
        ans++;
      }
    }
  }
  n -= p;
  u = 1;
  for (int i = 1; i < log10(n); i++) {
    u *= 2;
    ans += u;
  }
  if (n > 0) {
    checker(0, 0);
  }
  cout << ans << endl;
  ;
}
