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.math.BigInteger;
  import java.util.Scanner;
  
  public class Main {
      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();
              String k_2 = Integer.toBinaryString(k);
              BigInteger n_big = new BigInteger(n+"");
              BigInteger ans = new BigInteger("1");
              for (int i = 0; i < k_2.length(); i++) {
                  String str = k_2.substring(k_2.length()-i-1,k_2.length()-i);
                  BigInteger tmp = new BigInteger(str).multiply(n_big.pow(i));
                  ans = ans.add(tmp);
              }
              System.out.println(ans.mod(new BigInteger("1000000007")).subtract(BigInteger.ONE));
          }
      }
  }
  