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.*;
  import java.io.*;
  
  public class B {
      static MyScanner sc;
      static PrintWriter out;
      static {
          sc = new MyScanner();
          out = new PrintWriter(System.out);
      }
      public static void solve() {
          int n = sc.nextInt();
          char[] s = Integer.toString(sc.nextInt(), 2).toCharArray();
          long ans = 0, pow = 1;
          for(int i = s.length - 1; i >= 0; i--) {
              if(s[i] == '1') ans += pow;
              pow *= n;
              pow %= 1000000007;
              ans %= 1000000007;
          }
          out.println(ans);
      }
      public static void main(String[] args) {
          int t = sc.nextInt();
          while(t-- > 0) solve();
          out.flush();
      }
  }
  
  class MyScanner {
      BufferedReader br;
      StringTokenizer tok;
      MyScanner() {
          try { br = new BufferedReader(new InputStreamReader(System.in)); }
          catch(Exception e) { System.out.println(e); }
          tok = new StringTokenizer("");
      }
      public String next() {
          try {
              while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
          }
          catch(Exception e) { System.out.println(e); }
          return tok.nextToken();
      }
      public int nextInt() {
          return Integer.parseInt(next());
      }
      public long nextLong() {
          return Long.parseLong(next());
      }
      public double nextDouble() {
          return Double.parseDouble(next());
      }
  }
  