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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  
  /**
   * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
   */
  
  public class Main {
     public static void main(String[] args) throws IOException {
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(in.readLine());
        for(int test=0;test<t;test++){
           String s[]=in.readLine().split(" ");
           Long n=Long.parseLong(s[0]);
           int k=Integer.parseInt(s[1]);
           boolean bits[]=new boolean[70];
           Long base=1l;
           int i=0;
           while(k>0){
              if(k%2==1) bits[i]=true;
              k=k/2;
              i++;
           }
           Long result=0l;
           for(int j=0;j<i;++j){
              if(bits[j]==true) result+=base;
              result=result%1000000007;
              base=(base*n)%1000000007;
           }
           System.out.println(result);
        }
     }
  }