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.*;
  
  public class Main {
      static final long MOD = 1_000_000_007;
      public static long power(long x, long st){
          long res=1;
          for(long i=1; i<=st; i++){
              res*=x;
              res%=MOD;
          }
          return res;
      }
  
      public static void solve(Scanner in) {
          long n, k;
          n = in.nextInt();
          k = in.nextInt();
          long sum = 0;
          long cnt=0;
          while(k>0){
              sum = sum + (k % 2)*power(n, cnt)%MOD;
              sum %= MOD;
              cnt++;
              k/=2;
          }
          System.out.println(sum);
      }
  
  
      public static void main(String[] args) {
          int t;
          Scanner in = new Scanner(System.in);
          t = in.nextInt();
          while(t-->0){
              solve(in);
          }
  
      }
  }
  
  