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 permutation[100];
long long fact[100];
void writeFact(int n) {
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i;
}
void kthPermutation(int n, int i) {
  i--;
  int j, k = 0;
  for (k = 0; k < n; ++k) {
    permutation[k] = i / fact[n - 1 - k];
    i = i % fact[n - 1 - k];
  }
  for (k = n - 1; k > 0; k--)
    for (j = k - 1; j >= 0; j--)
      if (permutation[j] <= permutation[k]) permutation[k]++;
  for (i = 0; i < n; i++) permutation[i]++;
}
long long lucky[100000];
void findLucky(int n) {
  lucky[0] = 4;
  lucky[1] = 7;
  int i = 0, now = 2;
  while (lucky[i] < n) {
    lucky[now] = lucky[i] * 10 + 4;
    now++;
    lucky[now] = lucky[i] * 10 + 7;
    now++;
    i++;
  }
}
int aksFact(int k) {
  int i = 0;
  while (k > fact[i]) i++;
  return i;
}
int sumLucky(int n) {
  int i = 0;
  while (lucky[i] <= n) i++;
  return i;
}
bool isLucky(int n, int m) {
  while (n > 0) {
    if (n % 10 != 4 && n % 10 != 7) return false;
    n /= 10;
  }
  while (m > 0) {
    if (m % 10 != 4 && m % 10 != 7) return false;
    m /= 10;
  }
  return true;
}
int main() {
  writeFact(15);
  int n, k;
  cin >> n >> k;
  if (n < 14 && k > fact[n]) {
    cout << -1 << endl;
    return 0;
  }
  findLucky(n + 100);
  int a = aksFact(k);
  int ans = sumLucky(n - a);
  kthPermutation(a, k);
  for (int i = 0; i < a; i++) {
    permutation[i] += n - a;
    ans += isLucky(permutation[i], n - a + i + 1);
  }
  cout << ans << endl;
  return 0;
}
