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 Codeforces {
  	static long mod = 1000000007;
  	static long solve(long a, long b) 
  	{ 
  	    long temp; 
  	    if( b == 0) 
  	        return 1l; 
  	    temp = solve(a, b / 2); 
  	    if (b % 2 == 0) 
  	        return (temp%mod*temp%mod)%mod; 
  	    else
  	        return (a%mod*temp%mod*temp%mod)%mod; 
  	} 	
  
  
  	public static void main(String[] args) {
  		Scanner sc = new Scanner(System.in);
  		while(sc.hasNext()) {
    int t = sc.nextInt();
    while(t-->0) {
  	long n = sc.nextLong();
  	long k = sc.nextLong();
  	String s = Long.toBinaryString(k);
  	long ans = 0;
  
    long cnt = 0;
  //	System.out.println(s);
    for(int i = s.length() - 1 ; i>=0 ; i--) {
  	
  		if(s.charAt(i) == '1') {
  			ans= (long) (ans%mod + solve(n,cnt)%mod)%mod;
  		}
  		cnt++;
  	
    }
  	System.out.println(ans%mod);
  		
  		}
  	}
  }
  }