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: /* Author: Luna
   * Date: 08-October-2021
   * Time: 21:11:34
   * Problem Link: 'https://codeforces.com/contest/1594/problem/B',
  */
  import java.util.*;
  import java.io.*;
  public class B_Special_Numbers
  {
      public static void main(String args[])
      {
          Scanner in = new Scanner(System.in);
          int t = in.nextInt();
          while(t-->0)
          {
              solve(in);
          }
          in.close();
      }
  
      private static void solve(Scanner in) {
          int n = in.nextInt();
          int k = in.nextInt();
          // if(k==1)
          // {
          //     System.out.println(1);
          //     return;
          // }
          int ans = 0;
          int MOD = 1_000_000_000 + 7;
  
  
  
          // Check for set bits in k. And the raise n to the position of these set bits
          for(int i=0;i<=31;i++)
          {
              if(((1<<i)&k)>0)
              {
                  // System.out.println("At "+i);
                  ans = (ans%MOD + pow(n, i, MOD))%MOD;
              }
          }
          System.out.println(ans);
  
  
              
  
      }
  
      private static int pow(int n, int i, int MOD) {
          int ans = 1;
          for(int j=1;j<=i;j++)
          {
              ans = mulmod(n, ans, MOD);
              // System.out.println(ans);
          }
          return ans%MOD;
      }
  
      static int mulmod(int a, int b,
                        int mod) {
          int res = 0; // Initialize result
          a = a % mod;
          while (b > 0) {
              if (b % 2 == 1) {
                  res = (res + a) % mod;
              }
  
              // Multiply 'a' with 2
              a = (a * 2) % mod;
  
              // Divide b by 2
              b /= 2;
          }
  
          // Return result
          return res % mod;
      }
  }