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;
  const int MS = 1000001;
  const long long int INF = 1e9 + 7;
  void usaco(string filename) {
    string in = filename + ".in", out = filename + ".out";
    freopen(in.c_str(), "r", stdin);
    freopen(out.c_str(), "w", stdout);
  }
  template <class T>
  bool comp(T a, T b) {
    if (a < b) return true;
    return false;
  }
  long long int po(long long int b, long long int p) {
    long long int ans = 1;
    while (p) {
      if (p % 2 == 0) {
        b = (b * b) % INF;
        p = p / 2;
      } else {
        ans = (ans * b) % INF;
        p--;
      }
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      long long int ans = 0;
      long long int cnt = 0, her = log2(k);
      while (cnt <= her + 1) {
        if (k & (1 << cnt)) ans += po(n, cnt);
        cnt++;
        ans %= INF;
      }
      cout << ans % INF << "\n";
    }
  }
  