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 NewClass
  {
     
   public static  void main(String args[])
     {
           int t,n,k;
        Scanner in=new Scanner (System.in);
        StringBuffer sb=new StringBuffer();
        t=in.nextInt();
        while(t-->0)
        {
            long p=1,ans=0;
        n=in.nextInt();
        k=in.nextInt();
       for(int i=0;i<32;i++)
        {
            if((k>>i&1)==1)
            {
                ans=(ans+p)%1000000007;
            }
            p=(p*n)%1000000007;
                
         
        }
        sb.append(ans%1000000007+"\n");
        }
         System.out.println(sb);
     
     }
  
   
  }
  