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 MAXF = 20;
long long f[MAXF];
int a[MAXF];
int p[MAXF];
bool isLucky(long long x) {
  do
    if (x % 10 != 4 && x % 10 != 7) return false;
  while (x /= 10);
  return true;
}
long long count(long long n) {
  long long res = 0;
  ;
  for (int i = 1; i <= 10; i++) {
    for (int j = 0; j < (1 << i); j++) {
      long long x = 0;
      for (int k = i - 1; k >= 0; k--) {
        (x *= 10) += 4;
        if (j & (1 << k)) x += 3;
      }
      if (x > n) return res;
      res++;
    }
  }
  return res;
}
int main() {
  long long n, m, k;
  scanf("%I64d %I64d", &n, &k);
  f[0] = 1;
  for (int i = 1; i < MAXF; i++) f[i] = f[i - 1] * i;
  for (m = 0; k > f[m]; m++)
    ;
  if (m > n) {
    puts("-1");
    return 0;
  }
  n -= m;
  long long ans = count(n);
  k--;
  for (int i = 0; i < m; i++) a[i] = i;
  for (int i = 0; i < m; i++) {
    int t = k / f[m - i - 1];
    k %= f[m - i - 1];
    p[i] = a[t];
    for (int j = t; j < m - i; j++) a[j] = a[j + 1];
  }
  for (int i = 0; i < m; i++) {
    if (isLucky(n + i + 1) && isLucky(n + p[i] + 1)) ans++;
  }
  printf("%I64d\n", ans);
  fprintf(stderr, "Time of execution: %.3lf sec.\n",
          (double)clock() / CLOCKS_PER_SEC);
  return 0;
}
