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>
int lc[2000], nl;
int ex(int a) {
  int i;
  for (i = 0; i < nl; i++)
    if (a == lc[i]) return 1;
  return 0;
}
int main() {
  int arr[20];
  int cr[20];
  int i, j, l;
  int n, k, m;
  long long f[20];
  int ans = 0;
  f[0] = f[1] = 1;
  for (i = 2; f[i - 1] < 1000000000; f[i] = f[i - 1] * i, i++)
    ;
  for (i = 0; i < 2000; i++) lc[i] = 0;
  nl = 0;
  for (i = 1; i <= 9; i++)
    for (k = 0; k < (1 << i); k++, nl++)
      for (j = 1; j < (1 << i); j <<= 1) {
        lc[nl] *= 10;
        lc[nl] += ((k & j) ? 7 : 4);
      }
  scanf("%d %d", &n, &k);
  for (i = 1; f[i] < k; i++)
    ;
  m = i;
  for (i = 0; i < nl and lc[i] <= (n - m); ans++, i++)
    ;
  for (i = 0; i < m; i++) arr[i] = n - m + i + 1;
  for (i = 0; i < m; i++) {
    j = k / (f[m - i - 1]) - (k % f[m - i - 1] == 0);
    cr[i] = arr[j];
    k -= f[m - i - 1] * j;
    for (; j < (m - i - 1); j++) arr[j] = arr[j + 1];
  }
  for (i = 0; i < m; i++) ans += (ex(n - m + i + 1) and ex(cr[i]));
  printf("%d\n", ans);
  return 0;
}
