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, i, j, k, offset, nn, sol;
int v[5500], nc[5500], fol[5500];
set<int> g;
long long fact[110], d[110][110];
int main() {
  scanf("%d%d", &n, &k);
  v[1] = 4;
  v[2] = 7;
  nc[1] = nc[2] = 1;
  nn = 2;
  for (int i = 1; i <= nn; ++i) {
    if (v[i] <= n - 13) ++sol;
    g.insert(v[i]);
    if (nc[i] == 9) continue;
    v[++nn] = v[i] * 10 + 4;
    nc[nn] = nc[i] + 1;
    v[++nn] = v[i] * 10 + 7;
    nc[nn] = nc[i] + 1;
  }
  if (n > 13) {
    offset = n - 13;
    n = 13;
  }
  d[1][1] = 1;
  fact[1] = 1;
  for (int i = 2; i <= n; ++i) {
    fact[i] = fact[i - 1] * i;
    for (int j = 1; j <= i; ++j) d[i][j] = j * fact[i - 1];
  }
  for (int i = n; i; --i) {
    int poz = 1;
    for (int j = 1; j <= n; ++j) {
      if (fol[j] == 1) continue;
      if (d[i][poz] >= k) {
        int nr = offset + j;
        int pz = offset + n - i + 1;
        if (g.find(nr) != g.end() && g.find(pz) != g.end()) ++sol;
        fol[j] = 1;
        break;
      }
      k -= d[i][poz];
      ++poz;
    }
  }
  printf("%d\n", sol);
  return 0;
}
