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 int power(long long int base, long long int n) {
    long long int res = 1;
    while (n != 0) {
      if (n % 2 == 1) {
        res = (res * base) % (1000000000 + 7);
        n--;
      } else {
        base = (base * base) % (1000000000 + 7);
        n = n / 2;
      }
    }
    return res % (1000000000 + 7);
  }
  int main() {
    long long int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      string s;
      while (k != 0) {
        if (k % 2 == 0)
          s += '0';
        else
          s += '1';
        k = k / 2;
      }
      long long int ans = 0;
      for (int i = 0; i < s.size(); i++) {
        if (s[i] == '1') ans += power(n, i);
        ans %= (1000000000 + 7);
      }
      cout << ans << endl;
    }
  }
  