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 mod = 1e9 + 7;
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int T;
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      long long ans = 0;
      vector<long long> rank(32, 0), pn(32, 0);
      rank[0] = 1, pn[0] = 1;
      for (int i = 1; i < 31; i++) {
        rank[i] = (1ll << i) + rank[i - 1];
        pn[i] = (pn[i - 1] * n) % mod;
      }
      while (k > 0) {
        int p = 0;
        for (int i = 0; i < 31; i++) {
          if (rank[i] >= k) {
            p = i;
            break;
          }
        }
        ans = (ans + pn[p]) % mod;
        k = k - rank[p - 1] - 1;
      }
      cout << ans << '\n';
    }
    return 0;
  }
  