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.*;
  import java.math.*;
  import java.lang.*;
  
  /**
   *
   * @author free
   */
  public class JavaApplication1 {
  
      /**
       * @param args the command line arguments
       */
      public static void main(String[] args)throws IOException {
          // TODO code application logic here
          BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer tk;
          long MOD= 1000000007;
          long t= Integer.parseInt(br.readLine());
           
          
          for(int i=0;i<t;i++)
          {
              tk=new StringTokenizer(br.readLine());
              long n=Long.parseLong(tk.nextToken());
              long k=Long.parseLong(tk.nextToken());
              
              BigInteger ans =BigInteger.ZERO;
              int count=0;
              while(k>0){
                  
                  if( (k&1)==1L ){
                     
                      BigInteger temp= BigInteger.valueOf(n).pow(count);
                      ans=ans.add(temp);
                      ans=ans.mod(BigInteger.valueOf(MOD));
                  }
                  count++;
                  k=k>>1;
              }
              
              ans=ans.mod(BigInteger.valueOf(MOD));
              System.out.println(ans);
              
          }
      }
      
  }
  
  
  