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;
#pragma comment(linker, "/STACK:255000000")
bool firstout = 1;
template <class T>
void minn(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void maxx(T &a, T b) {
  if (a < b) a = b;
}
int madd(int &a, int b) {
  a += b;
  if (a >= 1000000009) a -= 1000000009;
  return a;
}
int msub(int &a, int b) {
  a -= b;
  if (a < 0) a += 1000000009;
  return a;
}
int mmult(int &a, int b) { return a = (long long)a * b % 1000000009; }
int mdiv(long long a, long long b, long long m) {
  a = (a % 1000000009 + 1000000009) % 1000000009;
  b = (b % 1000000009 + 1000000009) % 1000000009;
  if (a % b == 0) return a / b;
  return (a + m * mdiv(-a, m, b)) / b;
}
int n, m, l;
int A[10123];
int F[13];
set<int> s;
bool B[13];
bool happy(int a) { return s.count(a); }
int fnd(int n, int m, int p, int i = 0) {
  if (n == 0) return 0;
  int k = m / F[n - 1];
  int res = 0;
  int j;
  for (j = (0); j < (13); ++j) {
    if (!B[j]) --k;
    if (k < 0) break;
  }
  B[j] = 1;
  res = happy(p + i + 1) && happy(p + j + 1);
  return res + fnd(n - 1, m % F[n - 1], p, i + 1);
}
int slv(int n, int m, int p = 0) {
  --m;
  if (n < 13 && F[n] <= m) return -1;
  int i;
  for (i = (0); i < (13); ++i) B[i] = 0;
  return fnd(n, m, p);
}
int main() {
  int i, j, k;
  char c;
  int a, d;
  l = 0;
  for (i = (1); i < (10); ++i)
    for (j = (0); j < (1 << i); ++j) {
      int r = 0;
      int d = 1;
      for (k = (0); k < (i); ++k) r += d * (((j >> k) & 1) ? 7 : 4), d *= 10;
      A[l++] = r;
      s.insert(r);
    }
  F[0] = 1;
  for (i = (1); i < (13); ++i) F[i] = F[i - 1] * i;
  int ts;
  for (ts = 1; scanf("%"
                     "d",
                     &(n)) > 0;
       ++ts) {
    scanf(
        "%"
        "d",
        &(m));
    int res = 0;
    for (i = (0); i < (l); ++i)
      if (A[i] <= n - 13) ++res;
    if (n <= 13)
      res += slv(n, m);
    else
      res += slv(13, m, n - 13);
    printf(
        "%"
        "d",
        (res));
    printf("\n"), firstout = 1;
  }
  return 0;
}
