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;
  #pragma GCC optimize("O3", "unroll-loops")
  long long mod = 1e9 + 7;
  void solve();
  void tests() {
    long long test;
    cin >> test;
    while (test--) {
      solve();
    }
  }
  long long power(long long x, unsigned long long y, long long p) {
    long long res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  void io_files() {}
  int32_t main() {
    ;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    ;
    cout << fixed << setprecision(20);
    ;
    if (1) io_files();
    if (1)
      tests();
    else
      solve();
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    long long ans = 0;
    for (long long i = 0; i < 32; i++) {
      if ((1 << i) & k) ans += power(n, i, mod);
      ans %= mod;
    }
    cout << ans << endl;
  }
  