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 Solution {
  
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
            Scanner s =new Scanner(System.in);
       
           
          	  int t= s.nextInt();
        		for(int w=0;w<t;w++){
        		    long n=s.nextLong();
        		    long k=s.nextLong();
        		    long m=1;
        		    long ans=0;
        		    while(k!=0)
        		    {m=0;
        		    while(Math.pow(2,m)<=k)
        		    {
        		        m++;
        		    }
        		    
        		   
        		    m=m-1;
        		    ans=(ans+(long)modPow(n,m)%1000000007)%1000000007;
        		    
        		    k=k-(long)Math.pow(2,m);
        		    }
        		    System.out.println(ans);
            }
  	}
  	static long modPow(long var, long num) {
  	    long m = 1;
  	    while (num > 0) {
  	        m = (m * var) %1000000007;
  	        --num;
  	    }
  	    return m;
  
  }
  }