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 M = 1000000007;
  long long power(long long a, long long b) {
    long long res = 1;
    for (long long i = 0; i < b; i++) res = (res * a) % M;
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t, n, m, k, i, x, ara[100], ans;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      x = 0;
      m = k;
      ans = 0;
      while (m != 0) {
        ara[x] = m % 2;
        x++;
        m /= 2;
      }
      for (i = 0; i < x; i++) {
        if (ara[i] == 1) ans = ((ans % M) + power(n, i)) % M;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  