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 Round747B {
  
  	public static void main(String[] args)throws IOException {
  		// TODO Auto-generated method stub
  //		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  //		int t = Integer.parseInt(br.readLine().trim());
  //		StringTokenizer tk = new StringTokenizer(br.readLine());
  //		int m = Integer.parseInt(tk.nextToken(" "));
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(br.readLine().trim());
  		while(t-->0) {
  			StringTokenizer tk = new StringTokenizer(br.readLine());
  			int n = Integer.parseInt(tk.nextToken(" "));
  			int k = Integer.parseInt(tk.nextToken(" "));
  			long M = 1000000007;
  			long res = 0;
  			long p = 1;
  			while(k>0) {
  				int num = k%2;
  				k = k/2;
  				res = (res+num*p%M)%M;
  				p*=n;
  				p = p%M;
  				
  			}
  			System.out.println(res);
  			
  		}
  
  	}
  }