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 binpow(long long int a, long long int b, long long int m) {
    a %= m;
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  double log_base2(double d) { return log2(d); }
  double log_base2(double d);
  double findl(int k) { return log_base2(k); }
  double findl(int k);
  long long int find(int num, int kay) {
    long long int x;
    double y;
    int z;
    y = floor((findl(kay)));
    if (kay == 1) {
      x = 1;
    } else if (y == (findl(kay))) {
      x = binpow(num, y, 1000000007);
    } else {
      z = binpow(2, y, 1000000007);
      long long int b;
      b = binpow(num, y, 1000000007);
      long long int d;
      d = find(num, kay - z);
      if (d > 1000000007) {
        d = d % 1000000007;
      }
      x = b + d;
      if (x > 1000000007) {
        x = x % 1000000007;
      }
    }
    return x;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
      int n, k;
      cin >> n >> k;
      long long int ans;
      ans = find(n, k);
      cout << ans << endl;
    }
  }
  