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 long long int mod = 1e9 + 7;
  long long int binmod(long long int k, long long int x) {
    if (x == 0) return 1;
    if (x == 1) return k % mod;
    long long int res = binmod(k, x / 2);
    if (x % 2) return ((res * res % mod) * k) % mod;
    return res * res % mod;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int TESTCASE;
    cin >> TESTCASE;
    while (TESTCASE--) {
      long long int n, k;
      cin >> n >> k;
      vector<long long int> bit(32);
      for (long long int i = 0; i < 32; i++) {
        if (k & (1 << i)) {
          bit[i] = 1;
        }
      }
      long long int ans = 0;
      for (long long int i = 0; i < 32; i++) {
        if (bit[i]) ans = (ans + binmod(n, i)) % mod;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  