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 MAXL = 21;
int n, k;
int seq[MAXL], bound;
int used[MAXL];
void seq_cal() {
  bound = 1;
  long long tot = 1;
  for (; tot < k; ++bound, tot = tot * bound)
    ;
  memset(used, 0, sizeof(used));
  for (int num, i = bound; i > 0; --i) {
    tot /= i;
    seq[i] = (k - 1) / tot + 1;
    k = (k - 1) % tot + 1;
    num = seq[i];
    for (int j = 1; j <= bound; ++j)
      if (!used[j] && --num == 0) {
        used[j] = 1;
        seq[i] = (long long)j + n - bound;
      }
  }
}
int check_num(long long num) {
  num = n - num + 1;
  if (num <= bound)
    num = seq[num];
  else
    num = n - num + 1;
  for (; num; num /= 10)
    if (num % 10 != 4 && num % 10 != 7) return 0;
  return 1;
}
void solve() {
  seq_cal();
  int tot = 0;
  long long cur;
  for (int i = 1; i <= 10; ++i)
    for (int limit = (1 << i), j = 0; j < limit; ++j) {
      cur = 0;
      for (int l = i - 1; l >= 0; --l) {
        cur *= 10;
        if (j & (1 << l))
          cur += 7;
        else
          cur += 4;
      }
      if (cur > n) {
        printf("%d\n", tot);
        return;
      }
      if (check_num(cur)) ++tot;
    }
}
int main() {
  scanf("%d%d", &n, &k);
  solve();
  return 0;
}
