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 int BOUND = 1e9;
vector<int> lucky;
bool is_lucky(int x) {
  while (x > 0) {
    int t = x % 10;
    if (t != 4 && t != 7) {
      return false;
    }
    x /= 10;
  }
  return true;
}
void gen(int x) {
  if (x > BOUND) {
    return;
  }
  lucky.push_back(x);
  if (10ll * x + 4 <= BOUND) {
    gen(10 * x + 4);
  }
  if (10ll * x + 7 <= BOUND) {
    gen(10 * x + 7);
  }
}
int ans;
bool used[20];
long long fact[20];
int perm[20];
void small(int n, int k) {
  int ans = 0;
  int placed = 0;
  int need = n;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = fact[i - 1] * 1ll * i;
  }
  if (k > fact[n]) {
    puts("-1");
    exit(0);
  }
  for (int i = 0; i < n; i++) {
    long long cnt = 0;
    for (int j = 0; j < need; j++) {
      if (used[j]) {
        continue;
      }
      cnt++;
      if (cnt * fact[need - placed - 1] >= k) {
        perm[i] = 1 + j;
        if (is_lucky(1 + placed) && is_lucky(1 + j)) {
          ans++;
        }
        used[j] = true;
        k -= (cnt - 1) * fact[need - placed - 1];
        placed++;
        break;
      }
    }
  }
  printf("%d\n", ans);
  exit(0);
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  gen(4);
  gen(7);
  if (n <= 13) {
    small(n, k);
  }
  for (int i = 0; i < (int)lucky.size(); i++) {
    if (lucky[i] <= n - 13) {
      ans++;
    }
  }
  int placed = 0;
  int need = 13;
  fact[0] = 1;
  for (int i = 1; i <= 13; i++) {
    fact[i] = fact[i - 1] * 1ll * i;
  }
  for (int i = 0; i < 13; i++) {
    long long cnt = 0;
    for (int j = 0; j < need; j++) {
      if (used[j]) {
        continue;
      }
      cnt++;
      if (cnt * fact[need - placed - 1] >= k) {
        perm[i] = n - 12 + j;
        if (is_lucky(n - 12 + placed) && is_lucky(n - 12 + j)) {
          ans++;
        }
        used[j] = true;
        k -= (cnt - 1) * fact[need - placed - 1];
        placed++;
        break;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
