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;
long long fact[14];
void initFact() {
  fact[0] = 1;
  for (int i = (int)(1), _m = (int)(14); i < _m; ++i) fact[i] = fact[i - 1] * i;
}
int digits[11], n;
int dp[11][2];
int rec(int indx, int smaller) {
  if (indx == n) return 1;
  int &ret = dp[indx][smaller];
  if (ret != -1) return ret;
  ret = rec(indx + 1, true);
  if (smaller || 4 <= digits[indx])
    ret += rec(indx + 1, smaller || (4 <= digits[indx]));
  if (smaller || 7 <= digits[indx])
    ret += rec(indx + 1, smaller || (7 <= digits[indx]));
  return ret;
}
int countAll(int num) {
  n = 0;
  while (num) digits[n++] = num % 10, num /= 10;
  reverse(digits, digits + n);
  memset(dp, -1, sizeof dp);
  return rec(0, 0) - 1;
}
bool lucky(int n) {
  while (n)
    if (n % 10 == 4 || n % 10 == 7)
      n /= 10;
    else
      return false;
  return true;
}
vector<int> getPerm(vector<int> numbers, int k) {
  if (k == 0 || !numbers.size()) return numbers;
  int soso = fact[numbers.size() - 1];
  int koko = numbers[(k / soso)];
  numbers.erase(numbers.begin() + (k / soso));
  vector<int> ret = getPerm(numbers, k % soso);
  ret.insert(ret.begin(), koko);
  return ret;
}
int main() {
  initFact();
  int res, start, n, k;
  vector<int> perm;
  while (cin >> n >> k) {
    start = 0;
    for (int i = (int)(1), _m = (int)(14); i < _m; ++i)
      if (fact[i] >= k) {
        start = i;
        break;
      }
    start = n - start + 1;
    res = countAll(start - 1);
    perm.clear();
    for (int i = (int)(start), _m = (int)(n + 1); i < _m; ++i)
      perm.push_back(i);
    perm = getPerm(perm, k - 1);
    for (int pos = (int)(start), _m = (int)(n + 1); pos < _m; ++pos) {
      if (lucky(pos) && lucky(perm[pos - start])) ++res;
    }
    cout << res << endl;
  }
}
