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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.*;
  
  public class B1594 {
  
      public static void main(String[] args) {
          FastScanner fs = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          
          int t = fs.nextInt();
          
          while (t-- > 0) {
              long n = fs.nextLong();
              int k = fs.nextInt();
              
              String bin = toBin(k);
              long ans = 0;
              long base = 1;
              
              long MOD = 1000000007;
              for (int i = bin.length() - 1; i >= 0; i--) {
                  if (bin.charAt(i) == '1')
                      ans = (ans + base) % MOD;
                  base = (base * n) % MOD;
              }
              
              out.println(ans);
                  
          }
          
          out.close();
      }
  
      private static String toBin(int k) {
          String ans = "";
          
          while (k > 0) {
              ans = Integer.toString(k%2) + ans;
              k /= 2;
          }
          
          return ans;
      }
  
      static class FastScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
  }
  