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 dp[2502][102];
void intialise() {
  memset(dp, 0, sizeof dp);
  for (int i = 0; i <= 101; i++) dp[0][i] = 1;
  for (int j = 1; j <= 100; j++) {
    for (int i = 1; i <= 2500; i++) {
      for (int k = 0; k <= i && k <= 25; k++) {
        dp[i][j] = (dp[i][j] + dp[i - k][j - 1]) % 1000000007;
      }
    }
  }
}
int main() {
  intialise();
  int t, sz, sum;
  string s;
  cin >> t;
  while (t--) {
    cin >> s;
    sz = s.size();
    sum = 0;
    for (int i = 0; i < sz; i++) {
      sum += (s[i] - 'a');
    }
    cout << dp[sum][sz] - 1 << endl;
  }
  return 0;
}
