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 N = 1000000007;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      int n, k, i;
      cin >> n >> k;
      vector<int> v;
      v.push_back(1);
      for (i = 1; i < 30; i++) {
        v.push_back(((v[i - 1] % N) * (n % N)) % N);
        if (v[i] < 0) v[i] += N;
      }
      int binaryNum[30], b[30];
      for (i = 0; i < 30; i++) binaryNum[i] = 0;
      i = 0;
      while (k > 0) {
        binaryNum[i] = k % 2;
        k = k / 2;
        i++;
      }
      int ans = 0;
      for (i = 0; i < 30; i++) {
        ans = ((ans % N) + (binaryNum[i] * v[i]) % N) % N;
        if (ans < 0) ans += N;
      }
      cout << ans << endl;
    }
  }
  