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;
  void print(bool n) {
    if (n)
      cout << "YES";
    else
      cout << "NO";
    cout << '\n';
  }
  const long long MOD = 1000000007;
  long long add(long long x, long long y) {
    x += y;
    while (x >= MOD) x -= MOD;
    while (x < 0) x += MOD;
    return x;
  }
  long long sub(long long x, long long y) { return add(x, MOD - y); }
  long long mul(long long x, long long y) { return (x * 1ll * y) % MOD; }
  long long binpow(long long x, long long y) {
    long long z = 1;
    while (y > 0) {
      if (y % 2 == 1) z = mul(z, x);
      x = mul(x, x);
      y /= 2;
    }
    return z;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long ttt;
    cin >> ttt;
    for (long long T = 1; T <= ttt; T++) {
      long long n, k;
      cin >> n >> k;
      long long res = 0;
      for (long long i = 0; i < 35; i++) {
        if (k & (1ll << i)) {
          res = add(res, binpow(n, i));
        }
      }
      cout << res << '\n';
    }
    return 0;
  }
  