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 res;
int a[100], ins[100];
int IsLucky(int x) {
  for (; x; x /= 10)
    if (x % 10 != 4 && x % 10 != 7) return 0;
  return 1;
}
void Dfs(int now, int lim) {
  res++;
  if ((long long)now * 10 + 4 <= lim) Dfs(now * 10 + 4, lim);
  if ((long long)now * 10 + 7 <= lim) Dfs(now * 10 + 7, lim);
}
int main() {
  int n, m, k;
  scanf("%d%d", &n, &m);
  long long cnt = 1;
  for (int i = 1;; i++) {
    cnt *= i;
    if (cnt >= m) {
      k = i;
      break;
    }
  }
  if (k > n) {
    puts("-1");
    return 0;
  }
  memset(ins, 0, sizeof(ins));
  for (int i = 0; i < k; i++) {
    cnt = 1;
    for (int j = 1; j < k - i; j++) cnt *= j;
    int ret = (m - 1) / cnt;
    m -= ret * cnt;
    for (int j = 0; j < k; j++)
      if (!ins[j]) {
        if (ret == 0) {
          ins[j] = 1;
          a[i] = j;
          break;
        } else
          ret--;
      }
  }
  res = 0;
  for (int i = 0; i < k; i++)
    if (IsLucky(n - k + 1 + i) && IsLucky(n - k + 1 + a[i])) res++;
  if (4 <= n - k) Dfs(4, n - k);
  if (7 <= n - k) Dfs(7, n - k);
  printf("%d\n", res);
  return 0;
}
