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.*;
  import java.lang.*;
  import java.io.*;
  
  public class B_747_div2 {
  
  public static void main(String[] args) throws java.lang.Exception {
  	try {
  				
  	
  		
  		FastReader sc = new FastReader();
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			long n=sc.nextLong();
  			long k=sc.nextLong();
  			long res=0;
  			long p=1;
  			while(k>0)
  			{
  				if((k&1)>0)
  					res=(res+p)%1000000007;
  				p*=n;
  				p=p%1000000007;
  				k=k>>1;
  				}
  System.out.println(res);
  		}
  	} catch (Exception e) {
  	} finally {
  		return;
  	}
  }
  
  static class FastReader {
  	BufferedReader br;
  	StringTokenizer st;
  
  	public FastReader() {
  		br = new BufferedReader(new InputStreamReader(System.in));
  	}
  
  	String next() {
  		while (st == null || !st.hasMoreElements()) {
  			try {
  				st = new StringTokenizer(br.readLine());
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  		}
  		return st.nextToken();
  	}
  
  	int nextInt() {
  		return Integer.parseInt(next());
  	}
  
  	long nextLong() {
  		return Long.parseLong(next());
  	}
  
  	double nextDouble() {
  		return Double.parseDouble(next());
  	}
  
  	String nextLine() {
  		String str = "";
  		try {
  			str = br.readLine();
  		} catch (IOException e) {
  			e.printStackTrace();
  		}
  		return str;
  	}
  
  	int[] readArray(int n) {
  		int[] a = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nextInt();
  		return a;
  	}
  }
  
  }
  