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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
long long F[33];
vector<long long> v;
void go(int c, int len, int cur) {
  if (c > len) {
    v.push_back(cur);
    return;
  }
  go(c + 1, len, cur * 10 + 4);
  go(c + 1, len, cur * 10 + 7);
}
map<int, int> vis;
int lucky(int x) {
  while (x >= 1) {
    if (x % 10 != 4 and x % 10 != 7) return 0;
    x /= 10;
  }
  return 1;
}
int main() {
  F[0] = 1LL;
  for (int i = 1; i < 14; i++) F[i] = F[i - 1] * i;
  for (int i = 1; i <= 9; i++) go(1, i, 0);
  v.push_back(4444444444);
  int n, k;
  scanf("%d%d", &n, &k);
  k--;
  if (n <= 13 and F[n] < k) return printf("-1\n"), 0;
  int st = max(1, n - 13), ans = 0;
  for (int i = 0; i < int(v.size()); i++)
    if (v[i] < st) ans = i + 1;
  for (int i = st; i <= n; i++) {
    for (int j = st; j <= n; j++)
      if (!vis[j]) {
        if (k < F[n - i]) {
          ans += (lucky(i) & lucky(j));
          vis[j] = 1;
          break;
        }
        k -= F[n - i];
      }
  }
  printf("%d\n", ans);
  return 0;
}
