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.io.*;
  import java.util.*;
  public class Solution
  {
   
      public static void main(String args[])
      {
          Scanner sc=new Scanner(System.in);
          long T=sc.nextLong();
          for(long t=0;t<T;t++){
              long n=sc.nextLong();
              long k=sc.nextLong();
              String S=Long.toBinaryString(k);
  			
              long val=0;
              for(long i=0;i<S.length();i++){
                  long a;
  				if(S.charAt(S.length()-1-(int)i)=='0')
  					a=0;
  				else
  					a=1;
  				long b=1,m=1000000007;
  				for(int j=0;j<i;j++){
  					b=(b*n)%m;
  				}
                  
  				long c=a*b;
                  val=(val+c)%1000000007;
              }
  			
              System.out.println(val);
          }
      }
  }