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 m = 1e9 + 7;
  long long power(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a % m;
    if (b % 2 == 0) {
      return (power(a, b / 2) % m * power(a, b / 2) % m) % m;
    }
    long long x = (power(a, b / 2) % m * power(a, b / 2) % m) % m;
    return (a % m * x % m) % m;
  }
  int main() {
    long long t;
    cin >> t;
    vector<long long> v;
    long long x = 1;
    while (x < m) {
      v.push_back(x);
      x = x * 2;
    }
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<long long> res;
      while (k) {
        int r = lower_bound(v.begin(), v.end(), k) - v.begin();
        if (r == v.size())
          r--;
        else if (v[r] != k)
          r--;
        res.push_back(r);
        k -= v[r];
      }
      long long ans = 0;
      for (long long x : res) {
        ans = (ans % m + power(n, x) % m) % m;
      }
      cout << ans << endl;
    }
  }
  