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>
#pragma comment(linker, "/STACK:10000000")
const long double LDINF = 9128739847123.00;
const long double eps = 1e-9;
const int inf = 1 << 28;
const long double pi = fabsl(atan2l(0.0, -1.0));
using namespace std;
vector<long long> all;
long long n, k;
void load() {
  cin >> n >> k;
  {
    for (int i = 1; i <= 10; i++) {
      for (int j = 0; j < (1 << i); j++) {
        long long num = 0;
        for (int k = 0; k < i; k++) {
          if (j & (1 << k))
            num = num * 10 + 4;
          else
            num = num * 10 + 7;
        }
        all.push_back(num);
      }
    }
    sort(all.begin(), all.end());
  }
}
long long fact[14];
int count(int l, int r) {
  int pos1 = 0;
  int pos2 = 0;
  for (; pos1 < all.size() && all[pos1] < l; pos1++)
    ;
  for (; pos2 < all.size() && all[pos2] <= r; pos2++)
    ;
  return pos2 - pos1;
}
bool lucky(int a) {
  bool ok = true;
  while (a && ok) {
    if (a % 10 != 4 && a % 10 != 7) ok = false;
    a /= 10;
  }
  return ok;
}
long long a[1123];
void solve() {
  fact[0] = 1;
  for (int i = 1; i <= 13; i++) fact[i] = fact[i - 1] * i;
  if (n <= 13 && k > fact[n]) {
    printf("-1");
    return;
  }
  long long res = 0;
  int i = 1;
  int add = 0;
  int begin = 1;
  if (n - 1 > 13) {
    res += count(1, n - 14);
    add = n - 14;
    i = n - 13;
    begin = n - 13;
  }
  k--;
  int sz = 0;
  for (sz = 0; i <= n; sz++, i++) {
    a[sz] = k / fact[n - i];
    k %= fact[n - i];
  }
  for (int l = sz - 1; l >= 0; l--) {
    for (int j = l + 1; j < sz; j++) {
      if (a[j] >= a[l]) a[j]++;
    }
  }
  for (int i = 0; i < sz; i++)
    if (lucky(add + a[i] + 1) && lucky(begin + i)) res++;
  cout << res;
}
int main() {
  load();
  solve();
  return 0;
}
