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 n, K, j;
long long nums[100020];
long long f[16];
int a[16];
int us[16];
bool findN(int x) {
  for (int i = 1; i <= j; i++)
    if (nums[i] == x) return 1;
  return 0;
}
long long nex(long long x) {
  for (int i = 1; i <= j; i++) {
    if (nums[i] > x) return nums[i];
  }
  return nums[j];
}
int main() {
  int i, k, le;
  int ans = 0;
  scanf("%d %d", &n, &K);
  j = 0;
  for (le = 1; le < 10; le++) {
    for (k = 0; k < (1 << le); k++) {
      j++;
      nums[j] = 0;
      for (i = 0; i < le; i++) {
        if (k & (1 << i))
          nums[j] = nums[j] * 10 + 7;
        else
          nums[j] = nums[j] * 10 + 4;
      }
    }
  }
  j++;
  nums[j] = 4444444444LL;
  sort(nums + 1, nums + j + 1);
  ans = 0;
  for (i = 1; i <= j; i++) {
    if (nums[i] + 13 <= n) {
      ans++;
    }
  }
  f[0] = 1;
  us[0] = 1;
  for (i = 1; i <= 13; i++) {
    a[i] = i;
    f[i] = f[i - 1] * i;
    us[i] = 0;
  }
  K--;
  for (i = 1; i <= 13; i++) {
    k = 0;
    while ((long long)k * f[13 - i] <= K) k++;
    k--;
    K = K - f[13 - i] * k;
    for (int l = 1; l <= 13; l++) {
      if (us[l] == 0) {
        if (k == 0) {
          a[i] = l;
          us[l] = 1;
          break;
        }
        k--;
      }
    }
  }
  for (i = 1; i <= 13; i++) {
    a[i] = a[i] - 13 + n;
  }
  for (i = 1; i <= 13; i++) {
    if ((findN(a[i]) == 1) && (findN(i - 13 + n))) {
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
