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.Scanner;
  
  public class B {
      static int MOD = (int) 1e9 + 7;
  
      public static void main(String[] args) {
          Scanner scn = new Scanner(System.in);
          int t = scn.nextInt();
          StringBuilder sb= new StringBuilder();
          while (t-- > 0) {
              int n = scn.nextInt();
              int k = scn.nextInt();
              char[] bin = Integer.toBinaryString(k).toCharArray();
              long ans = 0;
              long pow = 1;
              for (int i = bin.length - 1; i >= 0; i--) {
                  int cof=bin[i] - '0';
                  ans = (ans + cof * pow) % MOD;
                  pow = (pow * n) % MOD;
              }
              sb.append(ans);
              sb.append("\n");
          }
          System.out.println(sb);
      }
  
  
  }
  