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.io.*;
  import java.util.*;
  
  public class SpecialNumbers {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  //        BufferedReader br = new BufferedReader(new FileReader("input.txt"));
          handleMultiTest(br);
      }
  
      private static void handleMultiTest(BufferedReader br) throws IOException {
          int T = Integer.parseInt(br.readLine());
          for (int i = 0; i < T; i++) {
              handleTest(br);
          }
      }
  
      private static void handleTest(BufferedReader br) throws IOException {
          long MOD = 1000000007;
          long[] nk = Arrays.stream(br.readLine().split(" ")).mapToLong(Long::parseLong).toArray();
          long n = nk[0];
          long k = nk[1];
          long result = 0;
          for (int i = 0; i <= 30; i++) {
              long isSetBit = (1 << i) & k;
              if (isSetBit > 0) {
                  result = (result + mathPow(n, i, MOD)) % MOD;
              }
          }
          System.out.println(result);
      }
  
      private static long mathPow(long ele, long pow, long MOD) {
          // System.out.println(ele + " ^ " + pow);
          if (pow == 0) return 1;
          else if (pow == 1) return ele % MOD;
          long res = 1;
          while (pow > 0) {
              res = (res * ele) % MOD;
              pow--;
          }
          // System.out.println(ele + " ^ " + pow + " %" + MOD + " = " + res);
          return res;
      }
  }