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 inf = 1 << 30;
const long long linf = 1LL << 60;
const int nlim = 20;
vector<int> lucks;
bool used[nlim];
void gen_lucky() {
  for (int len = 1; len <= 9; ++len) {
    for (int mask = 0; mask < (1 << len); ++mask) {
      int x = mask;
      int luck = 0;
      for (int j = 0; j < len; ++j) {
        luck *= 10;
        luck += (x % 2) ? 7 : 4;
        x /= 2;
      }
      lucks.push_back(luck);
    }
  }
  sort((lucks).begin(), (lucks).end());
}
long long fact(int n) {
  long long res = 1;
  for (int i = 2; i < n; ++i) {
    res *= i;
  }
  return res;
}
bool is_luck(int x) {
  for (int i = 0; i < (int)(lucks).size(); ++i) {
    if (lucks[i] == x) {
      return 1;
    }
  }
  return 0;
}
int getn(int n) {
  ++n;
  for (int i = 0; i < nlim; ++i) {
    if (!used[i]) {
      --n;
    }
    if (!n) {
      return i;
    }
  }
  return -1;
}
int main() {
  gen_lucky();
  int n, k;
  cin >> n >> k;
  int st = 1;
  if (n > nlim) {
    st = n - nlim + 1;
  }
  int ans = 0;
  for (int i = 0; i < (int)(lucks).size(); ++i) {
    if (lucks[i] < st) {
      ++ans;
    }
  }
  int pos = 0;
  while (k && st + pos <= n) {
    long long f = fact(n - st - pos + 1);
    long long cnt = k / f;
    if (k % f == 0) {
      --cnt;
    }
    cnt = min((long long)n - st - pos, cnt);
    int x = getn(cnt);
    used[x] = 1;
    if (is_luck(x + st) && is_luck(pos + st)) {
      ++ans;
    }
    ++pos;
    k -= cnt * f;
  }
  if (k > 1) {
    ans = -1;
  }
  cout << ans << "\n";
  return 0;
}
