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;
bool check(int n, int pos) { return (n & (1 << pos)) >> pos; }
const int M = 100010;
vector<int> lucky;
map<int, int> islucky;
long long f[M];
int solve(int actualpos, long long k, vector<int> numbers) {
  if (numbers.size() == 1)
    return (numbers[0] == actualpos && islucky[actualpos]);
  else {
    int i;
    for (i = 0; i < numbers.size(); i++) {
      if (k - f[numbers.size() - 1] <= 0) break;
      k -= f[numbers.size() - 1];
    }
    int rdx = ((numbers[i] == actualpos) && islucky[actualpos]);
    numbers.erase(numbers.begin() + i);
    return rdx + solve(actualpos + 1, k, numbers);
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, k, n;
  for (i = 1; i <= 9; i++) {
    for (j = 0; j < (1 << i); j++) {
      char a[15];
      for (k = 0; k < i; k++) a[k] = check(j, k) ? '4' : '7';
      a[i] = '\0';
      lucky.push_back(atoi(a));
    }
  }
  sort((lucky).begin(), (lucky).end());
  for (int a : lucky) islucky[a] = 1;
  scanf("%d%d", &n, &k);
  vector<int> imp;
  for (i = n; i > 0 && imp.size() < 14; i--) imp.push_back(i);
  sort((imp).begin(), (imp).end());
  f[0] = 1;
  for (i = 1; i < 17; i++) f[i] = i * f[i - 1];
  if (n < 15 && k > f[n]) {
    printf("%s\n", "-1");
    return 0;
  };
  int ls = lower_bound((lucky).begin(), (lucky).end(), imp[0]) - lucky.begin();
  int gr = solve(imp[0], k, imp);
  printf("%d\n", ls + gr);
  return 0;
}
