description: Theofanis really likes sequences of positive integers, thus his teacher (Yeltsa Kcir) gave him a problem about a sequence that consists of only special numbers.
  
  Let's call a positive number special if it can be written as a sum of different non-negative powers of n. For example, for n = 4 number 17 is special, because it can be written as 4^0 + 4^2 = 1 + 16 = 17, but 9 is not.
  
  Theofanis asks you to help him find the k-th special number if they are sorted in increasing order. Since this number may be too large, output it modulo 10^9+7.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first and only line of each test case contains two integers n and k (2 ≤ n ≤ 10^9; 1 ≤ k ≤ 10^9).
  
  Output
  
  For each test case, print one integer — the k-th special number in increasing order modulo 10^9+7.
  
  Example
  
  Input
  
  
  3
  3 4
  2 12
  105 564
  
  
  Output
  
  
  9
  12
  3595374
  
  Note
  
  For n = 3 the sequence is [1,3,4,9...]
solution: #include <bits/stdc++.h>
  using namespace std;
  long long rem = 1000000007;
  long long binaryexpo(long long x, long long n, long long m) {
    long long res = 1;
    while (n != 0) {
      if (n % 2 == 1) {
        res = ((res % m) * (x % m)) % m;
      }
      x = ((x % m) * (x % m)) % m;
      n /= 2;
    }
    return res;
  }
  int32_t main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      long long temp = log2(k);
      long long terms = pow(2, temp);
      if (k == terms) {
        long long var = binaryexpo(n, temp, rem);
        cout << var << "\n";
      } else {
        long long left = k - terms;
        long long ans = binaryexpo(n, temp, rem);
        while (left > 0) {
          long long go = log2(left);
          long long ter = pow(2, go);
          left -= ter;
          long long ad = binaryexpo(n, go, rem);
          ans = (ans + ad) % rem;
        }
        cout << ans << "\n";
      }
    }
  }
  