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 a[1 << 20];
long long b[30], c[30];
int num;
bool visit[40];
void dfs(int dep, long long now) {
  if (dep > 0) a[num++] = now;
  if (dep > 10) return;
  dfs(dep + 1, now * 10 + 4);
  dfs(dep + 1, now * 10 + 7);
}
bool check(long long x) {
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return 0;
    x /= 10;
  }
  return 1;
}
int main() {
  long long n, k;
  b[0] = 1;
  for (int i = 1; i <= 20; i++) b[i] = b[i - 1] * i;
  while (scanf("%I64d %I64d", &n, &k) == 2) {
    num = 0;
    dfs(0, 0);
    sort(a, a + num);
    long long sum = 1;
    bool flag = 0;
    for (int i = 1; i <= n; i++) {
      sum = sum * i;
      if (sum >= k) {
        flag = 1;
        break;
      }
    }
    if (!flag) {
      puts("-1");
      continue;
    }
    int p = 0;
    int ans = 0;
    long long res = max(1LL, n - 20) - 1;
    for (int i = 0; i < num; i++)
      if (a[i] <= res && check(i)) ans++;
    memset(visit, 0, sizeof(visit));
    for (int i = max(1LL, n - 20); i <= n; i++) {
      long long t = n - i;
      p++;
      long long t1 = (k + b[t] - 1) / b[t];
      k -= (t1 - 1) * b[t];
      for (int j = 1; j <= 20; j++)
        if (!visit[j]) {
          t1--;
          if (t1 == 0) {
            c[p] = j;
            visit[j] = 1;
            break;
          }
        }
    }
    for (int i = 1; i <= p; i++)
      if (check(res + c[i]) && check(res + i)) ans++;
    printf("%d\n", ans);
  }
  return 0;
}
