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.util.Scanner;
  public class SpecialNumber {
      public static void main(String[] args) throws Exception {
          Scanner r=new Scanner(System.in);
          int T=r.nextInt();
          while(T-->0) {
              int p = r.nextInt();
              int q = r.nextInt();
              GetAnswer(p, q);
          }
      }
      public static void GetAnswer (int n, int m){
          String s = Integer.toBinaryString(m);
          long res = 0;
          int num = s.length();
          long[] nums = new long[num];
          nums[0] = 1;
          for (int i = 1; i < num; i++) {
              nums[i] = (nums[i - 1]*n)%1000000007;
          }
          for (int j = 0; j < num; j++) {
              if (s.charAt(j) == '1') res = (res + nums[num - j - 1])%1000000007;
          }
          System.out.println(res);
      }
  }