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 A {
      static final Reader s = new Reader();
      static final PrintWriter out = new PrintWriter(System.out);
  
      public static void main(String[] args) throws IOException {
          int t  = s.nextInt();
  //        int t=1;
          for(int i=1; i<=t; ++i) {
  //            out.print("Case #"+i+": ");
              new Solver();
          }
          out.close();
      }
      static class Solver {
          static long M = (long)1e9+7;
          static long bin(long a, long b) {
              a%=M;
              long g = 1;
              while(b>0){
                  if((b&1)==1)g = g*a%M;
                  a = a*a%M;
                  b>>=1;
              }
              return g;
          }
          static int cnt(long n){
              int count = 0;
              int temp = 0;
              for(int i = 0 ; i <32 ; i++){
                  temp = 1;
                  temp = temp << i;
                  temp = (int) (n & temp);
                  if((temp > 0))
                      count++;
              }
              return count;
          }
          Solver() {
              long n = s.nextLong();
              long k = s.nextLong();
              String g = Long.toBinaryString(k);
              StringBuilder se =new StringBuilder(g);
              se = se.reverse();
              long v=0;
              for(int i=0;i<se.length();i++){
                 if(se.charAt(i)=='1')v = (v+(long)bin(n,i))%M;
              }
              out.println(v);
          }
      }
      static class Reader {
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st;
          String next() {
              while(st==null||!st.hasMoreTokens()) {
                  try {
                      st=new StringTokenizer(in.readLine());
                  } catch(Exception e) {}
              }
              return st.nextToken();
          }
          int nextInt() {
              return Integer.parseInt(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
  }