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;
const long long K = 10000000000LL;
vector<long long> L;
void init(long long crr) {
  if (crr > K) return;
  if (crr != 0) L.push_back(crr);
  init(crr * 10 + 4);
  init(crr * 10 + 7);
}
bool is_lucky(long long x) {
  if (x == 0) return false;
  while (x > 0) {
    if (x % 10 != 4 && x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
const int MAX_M = 20;
long long fact(long long k) {
  long long res = 1;
  for (int i = 1; i <= k; ++i) res *= i;
  return res;
}
vector<int> kth_permutation(int n, long long k) {
  vector<int> res(n), num(n);
  for (int i = 0; i < (int)(n); i++) num[i] = i;
  if (k >= fact(n)) return vector<int>();
  for (int i = 0; i < (int)(n); i++) {
    long long f = fact(n - i - 1);
    res[i] = num[k / f];
    num.erase(num.begin() + (k / f));
    k %= f;
  }
  return res;
}
long long perm_n, perm_m;
vector<int> perm_last;
bool init_permutation(long long n, long long k) {
  perm_n = n;
  perm_last = kth_permutation(min(n, 20LL), k);
  if (perm_last.empty()) return false;
  perm_m = perm_n - perm_last.size();
  return true;
}
long long get_permutation(long long i) {
  if (i < perm_m)
    return i;
  else
    return perm_m + perm_last[i - perm_m];
}
int bruteforce(int N, int K) {
  vector<int> A(N);
  for (int i = 0; i < (int)(N); i++) A[i] = i + 1;
  for (int i = 0; i < (int)(K - 1); i++)
    next_permutation((A).begin(), (A).end());
  int ans = 0;
  for (int i = 0; i < (int)(N); i++)
    if (is_lucky(i + 1) && is_lucky(A[i])) ++ans;
  return ans;
}
int main() {
  init(0);
  long long N, K;
  while (cin >> N >> K) {
    --K;
    if (!init_permutation(N, K)) {
      puts("-1");
      continue;
    }
    long long ans = 0;
    for (int i = 0; i < (int)(L.size()); i++) {
      long long l = L[i];
      if (1 <= l && l <= N && is_lucky(1 + get_permutation(l - 1))) ++ans;
    }
    cout << ans << endl;
  }
  return 0;
}
