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.*;
  
  public class Main{
      
      public static void main (String [] args){
          Scanner scan = new Scanner(System.in);
          int tc = scan.nextInt();
          for (int i = 0; i < tc; i++){
              long n = scan.nextLong();
              long k = scan.nextLong();
              System.out.println(solve(n,k));
          }
      }
  
      public static long solve (long n, long k){
          long mod = 1000000007;
          if (k == 1)
          return 1;
          //if (n == 2)
          //return k % mod;
          
          String s = Long.toBinaryString(k);
          //System.out.println(s);
          long result = 0;
          int len = s.length();
          long num = len-1;
          for (int i = 0; i < s.length(); i++){
              if (s.charAt(i) == '1')
              result += power(n,num);
              num--;
          }
          return result % mod;
      }
      
      public static long power(long n, long k){
          long result = 1;
          long mod = 1000000007;
          for (int i = 0; i < k; i++){
              result = ((result%mod)*(n%mod))%mod;
          }
          return result % mod;
      }
  }