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 fact[] = {1,    1,     2,      6,       24,       120,      720,
              5040, 40320, 362880, 3628800, 39916800, 479001600};
int perm[13];
int used[13];
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  if (a < 13 && b > fact[a]) {
    printf("-1\n");
    return 0;
  }
  int c = max(1, a - 12);
  int ret = 0;
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < (1 << i); j++) {
      int now = 0;
      for (int k = 0; k < i; k++) {
        now *= 10;
        if (j & (1 << k))
          now += 4;
        else
          now += 7;
      }
      if (now < c) ret++;
    }
  }
  int p = min(13, a);
  b--;
  for (int i = 0; i < p; i++) {
    int d = b / fact[p - i - 1];
    b %= fact[p - i - 1];
    int at = 0;
    int e = d + 1;
    while (1) {
      if (!used[at]) e--;
      if (e == 0) break;
      at++;
    }
    used[at] = 1;
    perm[i] = at;
  }
  for (int i = 0; i < p; i++) {
    int k = c + i;
    bool ok = true;
    while (k) {
      if (k % 10 != 4 && k % 10 != 7) ok = false;
      k /= 10;
    }
    k = c + perm[i];
    while (k) {
      if (k % 10 != 4 && k % 10 != 7) ok = false;
      k /= 10;
    }
    if (ok) ret++;
  }
  printf("%d\n", ret);
}
