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 mod = 1e9 + 7, n, m, ans, y, cnt;
  string s;
  short j;
  int inn() {
    int x;
    cin >> x;
    return x;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int _tc;
    cin >> _tc;
    while (_tc--) {
      n = inn();
      m = inn();
      ans = 0;
      j = 0;
      s = bitset<64>(m).to_string();
      reverse(s.begin(), s.end());
      long long h = 1;
      for (int i = 0; i < s.size(); i++) {
        h *= n;
        h %= mod;
        if (j == 0) h = 1;
        if (s[i] == '1') {
          ans += h % mod;
          ans %= mod;
        }
        j++;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  