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 LIM = (1E9 + 7);
  void calc(vector<long long> &pred, long long x, int c) {
    pred[0] = x;
    for (int i = 1; i < c; i++) {
      pred[i] = pred[i - 1] * pred[i - 1] % LIM;
    }
    return;
  }
  long long pow(vector<long long> &pred, long long x, long long n) {
    long long res = 0;
    int pos = 0;
    for (; n != 0; n = n >> 1) {
      if (n & 0x1 != 0) {
        long long cur = 1;
        for (int i = 0; pos >> i != 0; i++) {
          if (pos >> i & 0x1 != 0) {
            cur = cur * pred[i] % LIM;
          }
        }
        res = (res + cur) % LIM;
      }
      pos++;
    }
    return res;
  }
  int main() {
    int case_count;
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    cin >> case_count;
    for (int case_no = 0; case_no < case_count; case_no++) {
      long long b, k, c = 0;
      cin >> b >> k;
      for (long long tmp = k; tmp != 0; tmp = tmp >> 1) {
        c++;
      }
      vector<long long> pred(c);
      calc(pred, b, c);
      cout << pow(pred, b, k) << endl;
    }
    return 0;
  }
  