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.*;
  import java.math.BigInteger;
  public class Main {
      public static BigInteger power(long base,int exp){
          BigInteger bigBase=new BigInteger(Long.toString(base));
          return bigBase.pow(exp);
      }
      public static String KthSpecial(long base,long k){
          String number=Long.toBinaryString(k);
          BigInteger ans=new BigInteger("0");
          for(int i=number.length()-1;i>=0;i--){
              if(number.charAt(i)=='1')
                  ans=ans.add(power(base,number.length()-i-1));
          }
          ans=ans.mod(new BigInteger("1000000007"));
          return ans.toString();
      }
      public static void main(String[] args) {
          Scanner input=new Scanner(System.in);
          long tests=input.nextLong();
          ArrayList<String> answers=new ArrayList<>();
          for (long i = 0; i < tests; i++) {
              long n = input.nextLong();
              long k = input.nextLong();
              var temp=KthSpecial(n,k);
              answers.add(temp);
          }
          for (String answer : answers) {
              System.out.println(answer);
          }
      }
  }