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.*;
  import java.util.Scanner;
  
  
  public class Solution {
  	static int mod=1000000007;;
  	
  //	 static class Pair  implements Comparable<Pair>{
  //		 int a;
  //		 int b;
  //		 
  //		
  //	
  //		 
  //		 Pair(int tim,int index){  
  //			 this.b=index;
  //			  
  //			 this.a=tim;  
  //		
  //			 }  
  //		 public int compareTo(Pair st){  
  //			 if(a==st.a)  
  //			 return b-st.b;  
  //			 else if(a>st.a)  
  //			 return 1;  
  //			 else  
  //			 return -1;  
  //			 }  
  //	 } 
  //	 
  //	
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		Scanner sc=new Scanner(System.in);
  		int tt=sc.nextInt();
  		
  		
  		while(tt-->0){
  			
  			long n=sc.nextLong();
  			long k=sc.nextLong();
  			
  			
  			long ans=fun(n,k);
  		
  			System.out.println(ans);
  		
  		
  			
  		}
  		
  		
  		
  		
  		
  		
  	
  	}
  	 static long modulo(long a,long b,int n) {
  		 long res=1;
  		 
  		 while(b>0) {
  			 if((b&1)==1) {
  				 res=(res*(a%n))%n;
  			 }
  			 a=((a%n)*(a%n))%n;
  			 b=b>>1;
  			 
  		 }
  		 return res;
  	 }
  	
  	static long fun(long n,long k) {
  		
  		long var=1;
  		int pow=0;
  		while(var<=k) {
  			var*=2;
  			pow++;
  		}
  		
  		long ans=modulo(n,pow-1,mod);
  		
  		
  		
  		long rank=modulo(2l,pow-1,mod);
  		
  		
  //		System.out.println(rank+" "+pow+" "+ans);
  		if(rank!=k) {
  			
  			ans=(ans+fun(n,k-rank))%mod;
  		}
  		return ans;
  		
  		
  	}
  	
  	
  
  }
  