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: 
  import java.util.*;
  
  public class Main {
      public static Scanner input = new Scanner(System.in);
      public static List<Long> ans = new ArrayList<>();
      public static void main(String[] args) {
          int t = input.nextInt();
          while (t-- > 0) {
              solve();
          }
      }
  
      private static void solve() {
          long n = input.nextLong();
          long k = input.nextLong();
          long ans = 0;
          long MOD = (long) 1e9+7;
          for(long i=34 ; i>=0 ; i--){
              if(((k>>i)&1)>0){
                  ans = (ans%MOD + (power(n,i,MOD)%MOD))%MOD;
              }
          }
          System.out.println(ans%MOD);
  
  
  
  
  
  
  
  
  
      }
      public static long power(long x , long y , long mod){
          long res = 1;
  
          x = x % mod;
  
          while (y > 0)
          {
              long l = y & 1;
  
              if (l>0) {
                  res = ((res % mod) * (x % mod)) % mod;
              }
  
              y = y >> 1;
              x = ((x % mod) * (x % mod)) % mod;
          }
          return res;
      }
  }