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;
FILE *in;
FILE *out;
vector<long long> v;
void getNums(long long num) {
  if (num >= 10000000000LL) return;
  v.push_back(num);
  getNums(num * 10 + 4);
  getNums(num * 10 + 7);
}
int isLucky(int n) {
  int flag = 1;
  while (n) {
    if (n % 10 != 4 && n % 10 != 7) {
      flag = 0;
      break;
    }
    n /= 10;
  }
  return flag;
}
int dummy(int n, int k) {
  vector<int> v;
  for (int i = 1; i <= n; i++) v.push_back(i);
  for (int i = 1; i < k; i++) {
    if (i % 100 == 0) cout << "i is " << i << " out of " << k << endl;
    next_permutation(v.begin(), v.end());
  }
  for (int i = 0; i < (int)v.size(); i++)
    fprintf(out, "%d%c", v[i], i + 1 == (int)v.size() ? '\n' : ' ');
  int ans = 0;
  for (int i = 0; i < (int)v.size(); i++)
    if (isLucky(i + 1) && isLucky(v[i])) ans++;
  return ans;
}
int a[32];
int vis[32] = {0};
void encode(int idx, int n, int k) {
  if (idx >= n) return;
  long long perm = 1;
  for (int i = 1; i < n - idx; i++) perm *= i;
  for (int i = 0; i < n; i++)
    if (!vis[i]) {
      if (perm >= k) {
        vis[i] = 1;
        a[idx] = i;
        encode(idx + 1, n, k);
      } else
        k -= perm;
    }
}
int smarty(int n, int k) {
  int ans = 0, idx = 0;
  while (idx < (int)v.size() && v[idx] < (long long)n - 15) ans++, idx++;
  int pos = max(0, n - 15);
  encode(0, n - pos, k);
  for (int i = 0; i < n - pos; i++) a[i] += pos + 1;
  for (int i = 0; i < n - pos; i++)
    if (isLucky(i + pos + 1) && isLucky(a[i])) ans++;
  return ans;
}
int main(void) {
  in = stdin;
  out = stdout;
  getNums(4);
  getNums(7);
  sort(v.begin(), v.end());
  int n, k;
  fscanf(in, "%d %d", &n, &k);
  long long perm = 1;
  for (int i = 1; i <= n; i++) {
    perm *= i;
    if (perm > k) break;
  }
  if (perm < k) {
    fprintf(out, "-1\n");
    return 0;
  }
  fprintf(out, "%d\n", smarty(n, k));
  return 0;
}
