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 n, k, aa[10001], sum, cnt;
  void solve() {
    cin >> n >> k;
    aa[0] = 1;
    for (int i = 1; i <= 32; i++) {
      aa[i] = ((aa[i - 1] % m) * (n % m)) % m;
    }
    while (k) {
      if (k % 2 != 0) {
        sum = (aa[cnt] + sum % m) % m;
      }
      k /= 2;
      cnt++;
    }
    cout << sum << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    int t;
    cin >> t;
    while (t--) {
      solve();
      sum = cnt = 0;
      for (int i = 0; i < 36; i++) {
        aa[i] = 0;
      }
    }
  }
  