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 pariNa {
  
  	public static void main(String[] args) {
  		Scanner sc=new Scanner(System.in);
  		int t=sc.nextInt();
  		while(t-->0) {
  			int n=sc.nextInt();
  			int k=sc.nextInt();
  			long i=0,res=0,pres=0;
  			while(k>0) {
  				pres=(k%2)*power(n,i);
  				res=((res%1000000007)+(pres%1000000007))%1000000007;
  				k/=2;
  				i++;
  			}
  			System.out.println(res%1000000007);
  		}
  	}
  	static long power(long a,long b) {
  		long res=1;
  		if(a==0) {
  			return a;
  		}
  		if(b==0) {
  			return 1;
  		}
  		a=a%1000000007;
  		while(b>0) {
  			if((b&1)==1) {
  				res=(res*a)%1000000007;
  			}
  			a=(a*a)%1000000007;
  			b>>=1;
  		}
  		return res;
  	}
  }
  