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>
  #pragma GCC optimize("O3")
  using namespace std;
  const long long mod = 1e9 + 7;
  const long long base = 31;
  long long power(long long a, long long n, long long mod) {
    if (n == 0)
      return 1;
    else {
      long long temp = power(a, n / 2, mod) % mod;
      temp = (temp * temp) % mod;
      if (n % 2 == 1)
        return (a * temp) % mod;
      else
        return temp;
    }
  }
  void fre() {}
  clock_t start = clock();
  double timecomplexity;
  int n, k;
  long long ans = 0;
  void solve() {
    cin >> n >> k;
    ans = 0;
    for (int i = 0; i <= 31; i++)
      if (((k >> i) & 1) == 1) {
        ans += power(n, i, mod);
        ans %= mod;
      }
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
  }
  