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{
      public static void main(String[] args){
          Scanner s = new Scanner(System.in);
          int t= s.nextInt();
          for(int i=0;i<t;i++){
              int n=s.nextInt();
              int k =s.nextInt();
              int c=k;
              int a=0;
              for(;c!=0;){
                  c=c/2;
                  a++;
              }
              int A[]= new int[a];
              for(int j=0;j<a;j++){
                  A[j]=k%2;
                  k=k/2;
              }
              long mod= 1000000007;
              long ans=0;
              for(int j=0;j<a;j++){
                  if(A[j]==1){
                      long ans1=1;
                      for(int u=0;u<j;u++){
                          ans1= (ans1*n)%mod;
                      }
                      ans=(ans+ans1)%mod;
                  }
              }
              System.out.println(ans);
          }
      }    
  }