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 Solution {
      static long ans(int n, int k){
          String s=Integer.toBinaryString(k);
          long ans=0;
          long M=1000000007;
          int j=0;
          for(int i=0;i<s.length();i++){
              ans = ((n*ans)%M + (s.charAt(i)-'0'))%M;
          }
          return ans%M;
      }
      public static void main(String args[]){
          Scanner sc=new Scanner(System.in);
          int t=sc.nextInt();
          while(t-->0){
              int n=sc.nextInt();
              int k=sc.nextInt();
              if(k==1) System.out.println(1);
              else if(k==2) System.out.println(n);
              else if(k>2 && n==2) System.out.println(k);
              else{
                  System.out.println(ans(n, k));
              }
          }
          sc.close();
      }
  }
  