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.*;
  import java.util.*;
  
  public class Main {
    static long M = 1000000007;
    public static void debug(int[][] a){
      for(int i = 0 ; i < a.length ; i++){
        for (int j = 0 ; j < a[0].length ; j++){
          System.out.print(a[i][j] + " ");
        }
        System.out.println();
      }
    }
    public static long superPow(long a, long b){
      if(b == 0){
        return 1;
      }
  
      if(b % 2 == 0){
        return superPow(((a % M) * (a % M)) % M,b/2) % M;
      }
      else{
        return ((a % M) * (superPow(((a % M) * (a % M)) % M,(b-1)/2) % M)) % M;
      }
    }
    public static void main(String[] args) throws IOException {
      Scanner sc = new Scanner(System.in);
      int t = sc.nextInt();
  
      while (t-- > 0) {
        long n = sc.nextLong();
        long k = sc.nextLong();
  
        long mult = 0;
        long finalVal = 0;
  
        while(k != 0){
          long rem = k % 2;
          finalVal = (long) (((finalVal % M) + (((rem % M) * (superPow(n,mult) % M)) % M)) % M);
          k/=2;
          mult++;
        }
        System.out.println(finalVal);
      }
  
  
    }
  
    private static int getPrefVal(int i, int j, int k, int l, int[][] pref){
      return pref[k][l] + pref[i - 1][j - 1] - pref[i-1][l] - pref[k][j-1];
    }
  
    private static int getPref(int i, int j,int k, int l, int[][] pref) {
  
      int ans = 0;
  
      //outside - corner
      int count1 = getPrefVal(i+1,j,k-1,j,pref) + getPrefVal(i+1,l,k-1,l,pref) + getPrefVal(i,j+1,i,l-1,pref) + getPrefVal(k,j+1,k,l-1,pref);
      count1 = Math.abs((2*(k - i - 1 + l - j - 1)) - count1);
      //inner rect
      int count2 = 0;
      count2 = getPrefVal(i+1,j+1,k-1,l-1,pref);
      //count2 = Math.abs(((k-i-1) * (l-j-1)) - count2);
      return count1 + count2;
    }
  
  
  
  }
  