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 mod = (1e9) + 7;
const double eps = 1e-9;
const int siz = 1e5 + 5, siz2 = 1e5 + 5;
int n, k, ans;
vector<int> ve;
int fact(int v) { return v == 0 ? 1 : v * fact(v - 1); }
string cur;
bool check(int i) {
  string v = to_string(ve[i]);
  int sz = v.size();
  for (int i = 0; i < sz; i++) {
    if (v[i] != '7' && v[i] != '4') {
      return false;
    }
  }
  return true;
}
void solve() {
  long long curv = 0;
  if (cur != "") {
    curv = stoll(cur);
  }
  if (curv > n) {
    return;
  }
  if (curv >= n - min(13, n) + 1) {
    int index = curv - (n - min(13, n) + 1);
    ans += check(index);
  } else {
    ans += (curv != 0);
    cur += "4";
    solve();
    cur.pop_back();
    cur += "7";
    solve();
    cur.pop_back();
  }
}
int main() {
  scanf("%d%d", &n, &k);
  if (n < 13 && fact(n) < k) {
    printf("-1\n");
  } else {
    int v = n;
    for (int i = 0; i < min(n, 13); i++) {
      ve.push_back(v--);
    }
    reverse(ve.begin(), ve.end());
    int left = k - 1;
    int mn = min(12, n - 1);
    for (int i = 0; i < mn; i++) {
      int perm = fact(mn - i);
      for (int j = i + 1; j < min(n, 13); j++) {
        if (left - perm >= 0) {
          swap(ve[j], ve[i]);
          left -= perm;
        } else {
          break;
        }
      }
    }
    solve();
    printf("%d\n", ans);
  }
  return 0;
}
