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 i, j, k, n, m, an, l, r;
long long f[15], b[20], g[300000];
inline bool judge(int x, int y) {
  for (; x; x /= 10)
    if (!(x % 10 == 4 || x % 10 == 7)) return 0;
  for (x = y; x; x /= 10)
    if (!(x % 10 == 4 || x % 10 == 7)) return 0;
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  f[0] = 1;
  for (i = 1; i <= 14; i++) f[i] = f[i - 1] * i;
  if (n <= 14 && f[n] < m) {
    printf("-1\n");
    return 0;
  }
  int A = max(1, n - 14);
  k = A;
  g[l = r = 1] = 0;
  for (; l <= r; l++) {
    long long A = g[l] * 10 + 4;
    if (A < k) g[++r] = A;
    A += 3;
    if (A < k) g[++r] = A;
  }
  an = r - 1;
  for (i = A; i <= n; i++) {
    long long B = f[n - i], C = (m - 1) / B + 1;
    m -= (C - 1) * B;
    for (j = 1; C; j++)
      if (!b[j]) C--;
    b[j - 1] = 1;
    an += judge(i, A - 1 + j - 1);
  }
  printf("%d\n", an);
  return 0;
}
