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 ans, n, K, ar[15], l, st;
long long fac[15];
bool vis[15];
inline bool Check(int p) {
  while (p) {
    if ((p % 10 != 4) && (p % 10 != 7)) return false;
    p /= 10;
  }
  return true;
}
int main() {
  int i, j, k, p;
  long long tmp;
  scanf("%d%d", &n, &K);
  if (n < 13) {
    st = 0;
    l = n;
  } else {
    st = n - 12;
    l = 12;
    for (i = 1; i <= 10; ++i)
      for (j = (1 << i) - 1; j >= 0; --j) {
        tmp = 0;
        for (k = i - 1; k >= 0; --k) {
          tmp *= 10;
          if ((j >> k) & 1)
            tmp += 7;
          else
            tmp += 4;
        }
        if (tmp <= st) ++ans;
      }
  }
  fac[1] = 1;
  for (i = 2; i <= l; ++i) fac[i] = fac[i - 1] * i;
  for (--K, i = l; i > 1; --i) {
    p = K / fac[i - 1] + 1;
    K %= fac[i - 1];
    for (j = 1; j <= l; ++j) {
      if (!vis[j]) --p;
      if (!p) break;
    }
    ar[l - i + 1] = j;
    vis[j] = true;
  }
  for (i = 1; i <= l; ++i)
    if (!vis[i]) {
      ar[l] = i;
      break;
    }
  for (i = 1; i <= l; ++i)
    if (Check(st + ar[i]) && Check(st + i)) ++ans;
  printf("%d\n", ans);
  return 0;
}
