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.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class Special {
  	public static void main(String[] args) throws Exception{
  		BufferedReader buff = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer str = new StringTokenizer("");
  		StringBuilder string = new StringBuilder();
  		int cases = Integer.parseInt(buff.readLine());
  		for(int i = 0 ; i < cases ; i++) {
  			str = new StringTokenizer(buff.readLine());
  			long n = Long.parseLong(str.nextToken());
  			long k = Long.parseLong(str.nextToken());
  			long count = 0;
  			
  			while (k != 0) {
  				long log = (long) Math.floor(Math.log(k) / Math.log(2));
  				if(log == 0) {
  					count++;
  					k = k - 1;
  				} else {
  					count = (count + modPower(n , log)) % (Math.round(Math.pow(10, 9)) + 7);
  					k = k - modPower(2 , log);
  				}
  				
  			}
  			string.append(count + "\n");
  		}
  		
  		System.out.println(string);
  	}
  	
  	//doesnt work for b = 0
  	static long modPower (long a , long b) {
  		long mult = a;
  		for (int i = 0 ; i < b -1 ; i++) {
  			a = (a * mult) % (Math.round(Math.pow(10, 9)) + 7);
  		}
  		return a;
  	}
  }