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 maxn = 15;
int n, base = 0, K, ans, tmp, fac[maxn], used[maxn], res[maxn];
int calc(int n) {
  int res = 0;
  for (int t = 1; t <= 9; t++)
    for (int i = 0; i < (1 << t); i++) {
      int tmp = 0;
      for (int j = 0; j < t; j++)
        if (i >> j & 1)
          tmp = tmp * 10 + 4;
        else
          tmp = tmp * 10 + 7;
      res += tmp <= n;
    }
  return res;
}
bool check(int x) {
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
int main() {
  for (int i = 1; i <= 13; i++) fac[i] = fac[i - 1] * i;
  scanf("%d%d", &n, &K);
  if (n <= 13 && K > fac[n]) {
    puts("-1");
    return 0;
  }
  if (n >= 14) ans += calc(base = (n - 13));
  n -= base;
  fac[0] = 1;
  for (int i = n; i >= 1; i--) {
    tmp = (K - 1) / fac[i - 1];
    K -= tmp * fac[i - 1];
    for (int j = 1; j <= n; j++)
      if (!used[j] && !tmp) {
        used[j] = 1;
        res[n + 1 - i] = j;
        break;
      } else if (!used[j])
        tmp--;
  }
  for (int i = 1; i <= n; i++) ans += check(base + i) && check(base + res[i]);
  printf("%d\n", ans);
  return 0;
}
