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 int N = 1e5 + 5;
  const int M = 2e3 + 5;
  const int moh = 1e9 + 3;
  const long long mod = 1e9 + 7;
  const int base = 311;
  const int oi = 2e9;
  const long long oo = 1e18;
  const double eps = 1e-9;
  int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
  int dy[] = {-1, 0, 1, 0, -1, 1, -1, 1};
  int n, k;
  long long po[N];
  void solve() {
    cin >> n >> k;
    po[0] = 1;
    for (int i = 1; i <= 30; i++) {
      po[i] = po[i - 1] * n % mod;
    }
    long long ans = 0;
    for (int i = 30; i >= 0; i--) {
      if (k >> i & 1) {
        ans += po[i];
        ans %= mod;
      }
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int test;
    cin >> test;
    while (test--) {
      solve();
    }
  }
  