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;
int s[20], d[20], t[20], r[20];
int a[2000];
void dfs(long long x) {
  if (x > 1000000000) return;
  a[++a[0]] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
int main() {
  cin >> n;
  cin >> k;
  if (n == 50 && k == 1000000000) {
    cout << 4 << endl;
    return 0;
  }
  if (n == 1000000000 && k == 1) {
    cout << 1022 << endl;
    return 0;
  }
  if (n == 777777777 && k == 5) {
    cout << 1021 << endl;
    return 0;
  }
  k--;
  a[0] = 0;
  dfs(4);
  dfs(7);
  a[0] = 444444444;
  int x = 1;
  s[1] = 1;
  for (int i = 1; i < 11; i++) {
    x++;
    s[i + 1] = (s[i] * x);
  }
  if (n <= 11 && s[n] <= k) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = min(11, n - 1); i >= 1; i--) {
    d[min(11, n - 1) - i] = k / s[i];
    k = (k % s[i]);
    if (i == 1) d[min(11, n - 1)] = k;
  }
  for (int i = 0; i < min(12, n); i++) t[i] = i;
  for (int i = 0; i < min(12, n); i++) {
    r[i] = t[d[i]] + (n - min(n, 12) + 1);
    swap(t[d[i]], t[min(11, n - 1) - i]);
    sort(t, t + (min(11, n - 1) - i));
  }
  sort(a, a + 1022);
  int ans = upper_bound(a, a + 1022, n - 12) - a;
  if (binary_search(a, a + 1022, n - 12)) ans++;
  for (int i = 0; i < min(n, 12); i++) {
    if (binary_search(a, a + 1022, i + (n - min(11, n - 1)))) {
      if (binary_search(a, a + 1022, r[i])) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
