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 {
     // swap() doesn't swap i and j
    
     public static void main(String[] args) 
     {
         Scanner sc=new Scanner(System.in);
  		  int t=sc.nextInt();
  		  while(t-->0){
  		    int n=sc.nextInt();
  		    int k=sc.nextInt();
  		    int i=0;
  		    int l=1000000007;
  		    long sum=0;
  		    long p=1;
  		    
  		    for(i=0;i<31;i++){
  		        if((k & (1 << i))>0){
  		            sum=(sum%l+p%l)%l;
  		        }
  		        p=(p*n)%l;
  		        
  		    }
  		    System.out.println(sum);
  		    
  		    }
  		    
  		    
  		  
  		  
     }
  }