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;
const int MAX = 1000000000;
set<long long> st;
vector<long long> vec;
int N, K;
long long fac[20];
int num[20];
bool use[20];
void Init(long long x) {
  if (x < MAX) {
    st.insert(x * 10 + 4);
    Init(x * 10 + 4);
    st.insert(x * 10 + 7);
    Init(x * 10 + 7);
  }
}
void make_dp(int K, int m) {
  memset(use, 0, sizeof(use));
  for (int i = 1; i <= m; ++i) {
    int cnt = m - i;
    for (int j = m;; --j) {
      if (!use[j]) {
        if (cnt * fac[m - i] >= K) {
          --cnt;
        } else {
          use[j] = 1;
          num[i] = j;
          K -= cnt * fac[m - i];
          break;
        }
      }
    }
  }
}
bool luck_number(int x) {
  if (x <= 0) {
    return 0;
  }
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) {
      return 0;
    }
    x /= 10;
  }
  return 1;
}
int main() {
  Init(0);
  for (set<long long>::iterator it = st.begin(); it != st.end(); ++it) {
    vec.push_back(*it);
  }
  fac[1] = 1;
  for (int i = 2; i < 20; ++i) {
    fac[i] = fac[i - 1] * i;
  }
  while (cin >> N >> K) {
    if (N < 19 && fac[N] < K) {
      puts("-1");
      continue;
    }
    int m = 0;
    while (fac[m] < K) {
      m++;
    }
    N -= m;
    int ans = 0;
    for (int i = 0; vec[i] <= N; ++i) {
      ans++;
    }
    make_dp(K, m);
    for (int i = 1; i <= m; ++i) {
      if (luck_number(N + i) && luck_number(N + num[i])) {
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
