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.*;
  
  public class codeForces {
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          while (t-- > 0) {
              long n = sc.nextInt();
              long k = sc.nextInt();
              String s = Long.toBinaryString(k);
              long res = 0;
              s = String.valueOf(new StringBuilder(s).reverse());
              for (int i = 0; i < s.length(); i++) {
                  if (s.charAt(i) == '1') {
                      res = (res + pow(n, i, 1000000007));
                  }
              }
              System.out.println(res % 1000000007);
          }
      }
  
      public static long pow(long x, long y, long p) {
          long res = 1;
          x = x % p;
          if (x == 0)
              return 0;
          while (y > 0) {
              if ((y & 1) != 0)
                  res = (res * x) % p;
              y = y >> 1;
              x = (x * x) % p;
          }
          return res;
      }
  }
  