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 long long mod = 1000000007;
  const long long modx = 998244353;
  long double PI = 3.1415926535897;
  const long long N = 200010;
  long long pows(long long x1, long long y1) {
    long long r1 = 1;
    x1 = x1 % mod;
    while (y1 > 0) {
      if (y1 & 1) r1 = (r1 * x1) % mod;
      y1 = y1 >> 1;
      x1 = (x1 * x1) % mod;
    }
    return r1;
  }
  void Unstoppable_1();
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long test = 1;
    cin >> test;
    while (test--) {
      Unstoppable_1();
      cout << "\n";
    }
    return 0;
  }
  void Unstoppable_1() {
    long long n, k, ans = 0;
    cin >> n >> k;
    long long a[32], b[32];
    memset(b, 0, sizeof(b));
    for (long long i = 0; i < 32; i++) {
      a[i] = pows(n, i);
    }
    long long i = 0;
    while (k > 0) {
      b[i] = k % 2;
      k = k / 2;
      i++;
    }
    for (long long i = 0; i < 32; i++) {
      ans = (ans + (a[i] * b[i]) % mod) % mod;
    }
    cout << ans;
  }
  