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 yo {
  	public static void main(String[] args) {
  		StringBuilder str = new StringBuilder();
  		Scanner sc = new Scanner(System.in);
  		int t = sc.nextInt();
  		for (int xx = 0; xx < t; xx++) {
  			long n = sc.nextLong();
  			long k = sc.nextLong();
  			String k1 = Long.toBinaryString(k);
  
  			int c = 0;
  			long s = 0L;
  			for (int i = k1.length() - 1; i >= 0; i--) {
  				int c1 = 0;
  				if (k1.charAt(i) == '1')
  					c1 = 1;
  				Long k2 = 1L;
  				if (c1 == 1) {
  					for (int j = 0; j < c; j++)
  						k2 = (k2 * n) % 1000000007;
  				}
  				else
  				{
  					k2=0L;
  				}
  				s = (s + k2) % 1000000007;
  				c++;
  			}
  			System.out.println(s);
  		}
  	}
  }