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>
  const long long modulo = 1e9 + 7;
  using namespace std;
  long long binaryExponentiation(long long x, long long n, long long m) {
    if (n == 0) {
      return 1;
    }
    if (n % 2 == 0) {
      return binaryExponentiation(((x % m) * (x % m)) % m, n / 2, m);
    }
    return ((x % m) *
            binaryExponentiation(((x % m) * (x % m)) % m, (n - 1) / 2, m) % m) %
           m;
  }
  int32_t main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      long long i = 0, j = 1;
      while (k / j > 0) {
        j *= 2;
        i++;
      }
      long long s[i];
      for (long long p = 0; p < i; p++) {
        if (((1LL << p) & k)) {
          s[p] = 1;
        } else {
          s[p] = 0;
        }
      }
      long long ans = 0;
      for (long long p = 0; p < i; p++) {
        if (s[p] == 1) {
          ans += binaryExponentiation(n, p, modulo);
          ans %= modulo;
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  