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;
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      int arr[50], r = 0;
      long long count = 0;
      while (k != 0) {
        arr[r] = k % 2;
        k /= 2;
        r++;
      }
      for (int i = 0; i < r; i++) {
        long long k = 1;
        for (int j = i; j > 0; j--) {
          k *= n;
          k %= 1000000007;
        }
        count += arr[i] * k;
      }
      cout << count % (1000000007) << endl;
      for (int i = 0; i < 50; i++) arr[i] = 0;
    }
    return 0;
  }
  