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;
long long fact[44];
bool lucky(int n) {
  char cc[55];
  sprintf(cc, "%d", n);
  int m = strlen(cc);
  for (int i = 0; i < m; i++)
    if (cc[i] != '4' && cc[i] != '7') return 0;
  return 1;
}
int main() {
  int ret = 0;
  int N, K;
  fact[0] = 1;
  for (int i = 1; i < 15; i++) fact[i] = fact[i - 1] * i;
  scanf("%d %d", &N, &K);
  if (N < 14 && K > fact[N]) {
    printf("-1\n");
    return 0;
  }
  for (int l = 1; l <= 10; l++)
    for (int i = 0; i < (1 << l); i++) {
      long long n = 0;
      long long k = 1;
      for (int j = 0; j < l; j++) {
        if ((bool)(i & (1 << j)) == 0)
          n += k * 4;
        else
          n += k * 7;
        k *= 10;
      }
      if (n < N - 14 + 1) ret++;
    }
  int mask = 0;
  for (int i = max(0, N - 14); i < N; i++) {
    long long x = fact[N - max(0, N - 14) - (i - max(0, N - 14)) - 1];
    int c = 0;
    while (K > x) {
      K -= x;
      c++;
    }
    for (int j = 0; j < max(14, N - max(0, N - 14)); j++) {
      if (c == 0 && (mask & (1 << j)) == 0) {
        mask |= (1 << j);
        if (lucky(i + 1) && lucky(max(0, N - 14) + j + 1)) ret++;
        break;
      }
      if ((mask & (1 << j)) == 0) {
        c--;
      }
    }
  }
  printf("%d\n", ret);
  return 0;
}
