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 = 2000010;
int a[maxn], U, ans, vis[maxn];
long long fac[maxn];
void dfs(int pos, long long now) {
  if (now > U) return;
  if (now > 0) ans++;
  dfs(pos + 1, now * 10 + 4);
  dfs(pos + 1, now * 10 + 7);
}
bool check(int x) {
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
int main() {
  int N, K;
  fac[0] = 1;
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= 13; i++) fac[i] = fac[i - 1] * i;
  if (N <= 13 && fac[N] < K) return puts("-1"), 0;
  U = max(N - 13, 0);
  dfs(0, 0);
  for (int i = U + 1; i <= N; i++) {
    int t = 0, x = 0;
    while (fac[N - i] < K) K -= fac[N - i], t++;
    for (int j = 1; j <= N; j++) {
      if (vis[j]) continue;
      if (!t) {
        x = j;
        vis[x] = 1;
        break;
      }
      t--;
    }
    if (check(i) && check(U + x)) ans++;
  }
  printf("%d\n", ans);
  return 0;
}
