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 b {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          //BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
          StringTokenizer st = new StringTokenizer("");
          int mod = (int) 1e9 + 7;
  
          int tc = Integer.parseInt(br.readLine());
  
  
          for (int tt = 0; tt < tc; tt++) {
              st = new StringTokenizer(br.readLine());
              int factor = Integer.parseInt(st.nextToken());
              int k = Integer.parseInt(st.nextToken());
  
              long[] value = new long[32];
              long power = 1;
  
              for (int i = 0; i <= 31; i++) {
                  value[i] = power;
                  power *= factor;
                  power%=mod;
  
              }
  
              int answer = 0;
  
              for (int i = 0; k != 0; i++, k /= 2) {
                  if (k % 2 == 1) {
                      answer += value[i];
                      answer%=mod;
                  }
              }
  
              System.out.println(answer);
          }
      }
  
      private static int gcd(int a, int b) {
          if (a % b == 0) return b;
          else return gcd(b, a % b);
      }
  }