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 fact[(int)1000002];
  long long infac[(int)1000002];
  long long bin_pow(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a % (long long)1000000007;
    if (b % 2 == 0) {
      long long temp = bin_pow(a, b / 2);
      return (temp * temp) % (long long)1000000007;
    } else {
      return ((a % (long long)1000000007) * bin_pow(a, b - 1)) %
             (long long)1000000007;
    }
  }
  long long mul(long long a, long long b) {
    return ((a % (long long)1000000007) * (b % (long long)1000000007)) %
           (long long)1000000007;
  }
  void pre() {
    fact[0] = 1;
    for (long long i = 1; i < (int)1000002; i += 1)
      fact[i] = (i * fact[i - 1]) % (long long)1000000007;
    infac[(int)1000002 - 1] =
        bin_pow(fact[(int)1000002 - 1], (long long)1000000007 - 2);
    for (long long i = (int)1000002 - 1; i > 0; i -= 1)
      infac[i - 1] = (i * infac[i]) % (long long)1000000007;
  }
  long long nCr(long long n, long long r) {
    long long res = fact[n];
    res = (res * infac[n - r]) % (long long)1000000007;
    res = (res * infac[r]) % (long long)1000000007;
    return res;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    long long ans = 1;
    while (ans * 2 <= k) ans *= 2;
    long long idx;
    for (long long i = 63; i >= 0; i -= 1) {
      if ((ans & (1ll << i))) {
        idx = i;
        break;
      }
    }
    if (ans == k) {
      long long z = idx;
      long long res = bin_pow(n, z);
      cout << res << "\n";
    } else {
      k -= ans;
      long long res = bin_pow(n, idx);
      for (long long i = 63; i >= 0; i -= 1) {
        if ((k & (1ll << i))) {
          res += bin_pow(n, i);
        }
        res %= (long long)1000000007;
      }
      cout << res << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int tc;
    cin >> tc;
    while (tc--) {
      solve();
    }
    return 0;
  }
  