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;
long long ara[15], fact[15], ind[15];
bool chck(long long x) {
  vector<int> vec;
  while (x) vec.push_back(x % 10), x /= 10;
  for (int i = 0; i < vec.size(); i++)
    if (vec[i] != 4 && vec[i] != 7) return 0;
  return 1;
}
int perm(long long m, long long k) {
  for (int i = 1; i <= m - 1; i++) {
    long long f = fact[m - i];
    if (k <= f) continue;
    int cnt = 0;
    while (k > f) {
      k -= f;
      cnt++;
    }
    sort(ara + i - 1, ara + m);
    swap(ara[i - 1], ara[i + cnt - 1]);
  }
  int ret = 0;
  for (int i = 0; i < m; i++)
    if (chck(ind[i]) && chck(ara[i])) ret++;
  return ret;
}
char str[12];
int len, mem[12][2][2];
int dp(int pos, bool f, bool f2) {
  if (pos == len) return f2;
  if (mem[pos][f][f2] != -1) return mem[pos][f][f2];
  mem[pos][f][f2] = 0;
  if (!f2) mem[pos][f][f2] = dp(pos + 1, 1, f2);
  if (f)
    mem[pos][f][f2] += 2 * dp(pos + 1, f, 1);
  else {
    if (str[pos] >= '4') mem[pos][f][f2] += dp(pos + 1, str[pos] > '4', 1);
    if (str[pos] >= '7') mem[pos][f][f2] += dp(pos + 1, str[pos] > '7', 1);
  }
  return mem[pos][f][f2];
}
int func(long long n) {
  if (n == 0) return 0;
  while (n) str[len++] = n % 10 + '0', n /= 10;
  reverse(str, str + len);
  memset(mem, -1, sizeof mem);
  return dp(0, 0, 0);
}
int main() {
  fact[0] = 1;
  for (long long i = 1; i < 15; i++) fact[i] = (fact[i - 1] * i);
  long long n, k, s;
  scanf("%lld %lld", &n, &k);
  if (n < 13 && k > fact[n]) {
    puts("-1");
    return 0;
  }
  s = n - min(n, 13ll);
  int m = min(n, 13ll);
  for (int i = 0; i < m; i++) ara[i] = s + 1 + i, ind[i] = s + 1 + i;
  int res = func(s) + perm(m, k);
  printf("%d\n", res);
  return 0;
}
