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;
  int64_t mod = 1e9 + 7;
  int64_t log2(int64_t x, int64_t base) { return log(x) / log(base); }
  int64_t bin_pow(int64_t base, int64_t pow) {
    if (pow == 0) {
      return 1;
    }
    if (pow == 1) {
      return base;
    }
    int64_t res = bin_pow(base, pow / 2);
    res %= mod;
    res *= res;
    res %= mod;
    if (pow % 2 != 0) {
      res *= base;
      res %= mod;
    }
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int64_t t;
    cin >> t;
    while (t--) {
      int64_t n, k;
      cin >> n >> k;
      int64_t res = 0;
      while (k > 0) {
        int64_t n2p = log2(k, 2);
        res += bin_pow(n, n2p);
        res %= mod;
        k -= pow(2, n2p);
      }
      cout << res << "\n";
    }
  }
  