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;
int luck[5000], top;
int a[20], fa[20] = {1};
bool used[20];
void make(long long x) {
  if (x > 1e9) return;
  luck[top++] = x;
  make(x * 10 + 4);
  make(x * 10 + 7);
}
bool is(int x) { return x == 0 || (x % 10 == 4 || x % 10 == 7) && is(x / 10); }
void inttoarray(int k, int n) {
  int i, j, tmp;
  for (i = 1; i <= n; i++) {
    tmp = k / fa[n - i];
    for (j = 1; j <= n; j++)
      if (!used[j]) {
        if (tmp == 0) break;
        tmp--;
      }
    a[i] = j;
    used[j] = 1;
    k %= fa[n - i];
  }
}
int main() {
  int n, k, i, j, as = 0;
  for (i = 1; i < 20; i++) fa[i] = fa[i - 1] * i;
  scanf("%d%d", &n, &k);
  k--;
  if (n <= 13) {
    if (k >= fa[n]) return puts("-1");
    inttoarray(k, n);
    if (a[4] == 4 || a[4] == 7) as++;
    if (a[7] == 4 || a[7] == 7) as++;
    printf("%d\n", as);
    return 0;
  }
  make(0);
  sort(luck + 1, luck + top);
  inttoarray(k, 13);
  for (i = 1; i <= 13; i++)
    if (is(i + n - 13) && is(a[i] + n - 13)) as++;
  for (i = 1; i < top; i++)
    if (luck[i] > n - 13) break;
  as += i - 1;
  printf("%d\n", as);
  return 0;
}
