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 N, K;
vector<long long> L;
long long l, r;
long long fact[15];
void dfs(long long x, long long l, long long r) {
  if (x >= l && x <= r) L.push_back(x);
  if (x > r) return;
  dfs(x * 10 + 4, l, r);
  dfs(x * 10 + 7, l, r);
}
bool mark[20];
int perm[20];
bool lucky(long long x) {
  if (!x) return false;
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
int main() {
  fact[0] = 1;
  for (int i = 0; i < (int)(15); ++i)
    if (i) fact[i] = fact[i - 1] * i;
  cin >> N >> K;
  if (N < 13 && K > fact[N]) {
    printf("-1\n");
    return 0;
  }
  int pend = 1;
  while (K > fact[pend]) ++pend;
  for (int i = 0; i < (int)(pend); ++i) perm[i] = i;
  K--;
  for (int i = 0; i < pend; ++i) {
    int cnt = 0;
    for (int j = 0; j < (int)(pend); ++j)
      if (!mark[j]) {
        cnt++;
        if (cnt * fact[pend - i - 1] > K) {
          K -= (cnt - 1) * fact[pend - i - 1];
          perm[i] = j;
          mark[j] = true;
          break;
        }
      }
  }
  int ans = 0;
  for (int i = N - pend + 1; i <= N; ++i)
    if (lucky(i)) {
      int x = perm[i - (N - pend + 1)] + N - pend + 1;
      if (lucky(x)) ans++;
    }
  dfs(0, 1, N - pend);
  ans += L.size();
  cout << ans << endl;
  return 0;
}
