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.Scanner;
  import java.util.Arrays;
  
  public class Main {
      public static void main(String[] args) {
          Scanner scanner = new Scanner(System.in);
          int t = scanner.nextInt();
          for (int q = 0; q < t; q++) {
              long n= scanner.nextLong();
              long k= scanner.nextLong();
              long s=0;
              do{
                  long j= (long) (Math.log(k)/Math.log(2));
                  long b=(long) Math.pow(2,j);
                  k=k-b;
                  long r=func(n,j)%1000000007;
                  s=s+r;
              }while(k!=0);
              System.out.println(s%1000000007);
          }
      }
      public static long func(long a,long b){
          long p=1;
          for(int i=0;i<b;i++){
              p=p*a;
              p=p%1000000007;
          }
          return p;
      }
  }
  
  