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;
  const int mod = 1e9 + 7;
  long long gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long binopow(long long b, long long pow) {
    if (pow == 0) return 1;
    long long ans;
    ans = binopow(b, pow / 2);
    ans = (ans * ans) % mod;
    if (pow % 2) ans = (ans * b) % mod;
    return ans;
  }
  long long inv(long long a, long long mod) { return binopow(a, mod - 2); }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<int> v;
      while (k > 0) {
        v.push_back(k % 2);
        k = k / 2;
      }
      long long ans = 0;
      long long j = 1;
      for (int i = 0; i < v.size(); i++) {
        ans = ans + (v[i] * j) % mod;
        ans = ans % mod;
        j = (j * n) % mod;
      }
      if (ans < 0)
        cout << ans + mod;
      else
        cout << ans << "\n";
    }
  }
  