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;
long long P[2000], sz, Ans, N, K;
long long Q[10000];
int size;
void Dfs(int k, long long s) {
  Q[++size] = s;
  if (k == 11) {
    return;
  }
  Dfs(k + 1, s * 10 + 4);
  Dfs(k + 1, s * 10 + 7);
}
void Init() {
  Dfs(1, 0);
  sort(Q + 1, Q + size + 1);
}
long long Fac(long long s) {
  if (s == 1) return 1;
  return s * Fac(s - 1);
}
bool Check(long long n) {
  stringstream ss;
  string s;
  ss << n;
  ss >> s;
  bool f = 1;
  for (int i = 0; i < s.size(); i++)
    if (s[i] != '4' && s[i] != '7') {
      f = 0;
      break;
    }
  return f;
}
int main() {
  scanf("%I64d%I64d", &N, &K);
  K--;
  long long last = 1;
  Init();
  for (int i = 1; i <= size; i++)
    if (Q[i] < max(1ll, N - 13) && Check(Q[i])) Ans++;
  for (int i = max(1ll, N - 13); i < N; i++) {
    long long tmp = Fac(N - i);
    long long l = 0, r = N - i, mid, ans = 0;
    while (l <= r) {
      mid = (l + r) >> 1;
      if (mid * tmp > K)
        r = mid - 1;
      else
        ans = mid, l = mid + 1;
    }
    K -= ans * tmp;
    last += ans;
    ans++;
    for (int j = 1; j <= sz; j++)
      if (P[j] <= ans) ans++;
    P[++sz] = ans;
    sort(P + 1, P + sz + 1);
    if (Check(i) && Check(max(0ll, N - 14) + ans)) Ans++;
  }
  long long ans = 1;
  for (int j = 1; j <= sz; j++)
    if (P[j] <= ans) ans++;
  if (Check(N) && Check(max(0ll, N - 14) + ans)) Ans++;
  if (K > 0)
    puts("-1");
  else
    printf("%I64d\n", Ans);
}
