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 maxx = 1e9 + 7;
  long long power(int n, int r) {
    if (r == 1) {
      return n;
    } else if (r == 0) {
      return 1;
    }
    return (n * power(n, r - 1)) % maxx;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int a, b;
      cin >> a >> b;
      vector<int> binary;
      while (b > 0) {
        int sm = b % 2;
        b = b / 2;
        if (sm) {
          binary.push_back(1);
        } else {
          binary.push_back(0);
        }
      }
      long long sum = 0;
      for (int i = 0; i < binary.size(); i++) {
        if (binary[i]) {
          sum = (sum + power(a, i)) % maxx;
        }
      }
      cout << sum << endl;
    }
    return 0;
  }
  