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;
double PI = 3.141592653589793;
const int mx = 13;
long long cnt[mx + 1];
int val[mx + 1];
bool lck(int n) {
  bool out = 1;
  while (n != 0) {
    if (n % 10 != 4 && n % 10 != 7) out = 0;
    n /= 10;
  }
  return out;
}
const int bmx = 9;
vector<int> proc;
void init() {
  for (int i = 1; i < bmx + 1; i++) {
    for (int j = 0; j < (1 << i); j++) {
      int val = 0;
      for (int k = 0; k < i; k++) {
        if (j & (1 << k))
          val = 10 * val + 7;
        else
          val = 10 * val + 4;
      }
      proc.push_back(val);
    }
  }
}
void solve() {
  init();
  int n, k;
  cin >> n >> k;
  k--;
  cnt[0] = 1;
  for (int i = 1; i < mx + 1; i++) cnt[i] = cnt[i - 1] * i;
  if (n <= mx && k + 1 > cnt[n]) {
    cout << -1;
    return;
  }
  set<int> done;
  for (int i = 0; i < mx + 1; i++) {
    long long ccnt = 0, ind = 0;
    while (done.count(ind)) ind++;
    while (ccnt + cnt[mx - i] <= k) {
      ccnt += cnt[mx - i];
      ind++;
      while (done.count(ind)) ind++;
    }
    k -= ccnt;
    val[i] = ind;
    done.insert(ind);
  }
  int ans = 0;
  if (n <= mx + 1) {
    for (int i = mx - n + 1; i <= mx; i++) {
      int in = i - (mx - n);
      int v = val[i] - (mx - n);
      if (lck(in) && lck(v)) ans++;
    }
  } else {
    int st = n - mx;
    for (auto i : proc) ans += (i < st);
    for (int i = 0; i <= mx; i++) {
      int in = st + i;
      int v = st + (val[i]);
      if (lck(in) && lck(v)) ans++;
    }
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  for (int c = 1; c < T + 1; c++) {
    solve();
  }
}
