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 constexpr max_n = 1e5 + 10, max_lg = 30, mas_sq = 1000, mod = 1e9 + 7;
  long long bpow(long long x, long long y) {
    long long res = 1LL;
    while (y) {
      if (y & 1) {
        res *= x;
        res %= mod;
      }
      x *= x;
      x %= mod;
      y /= 2;
    }
    return res;
  }
  int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    long long t = 1;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      long long ans = 0LL;
      while (k > 0) {
        long long a = 0;
        for (long long i = 1; i <= k; i++) {
          if (((1LL << i) - 1LL) <= k) {
            a = i;
          } else
            break;
        }
        k -= ((1LL << a) - 1LL);
        if (k > 0) {
          ans += bpow(n, a);
          k--;
        } else {
          for (long long i = 0; i <= (a - 1); i++) {
            ans += bpow(n, i);
            ans %= mod;
          }
        }
        ans %= mod;
      }
      cout << (long long)ans << '\n';
    }
  }
  