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.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class Main {
      public static FastScanners fs = new FastScanners();
  
      public static void main(String[] args) {
          int testCase = 1;
          testCase = fs.nextInt();
          for (int i = 0; i < testCase; i++) {
              solve();
          }
      }
  
      public static void solve() {
          long n = fs.nextLong();
          long k = fs.nextLong();
          long ans = 0;
          long modulo = 1000000007;
          for (int i = 0; i <= 30; i++) {
              if ((k & (1 << i)) > 0) {
                  ans += power(n, i, modulo);
                  ans %= modulo;
              }
          }
          System.out.println(ans);
      }
  
      static long power(long x, long y, long p) {
          long res = 1;
          x = x % p;
  
          if (x == 0)
              return 0;
  
          while (y > 0) {
              if ((y & 1) != 0)
                  res = (res * x) % p;
  
              y = y >> 1; // y = y/2
              x = (x * x) % p;
          }
          return res;
      }
  
  }
  
  
  class FastScanners {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer st = new StringTokenizer("");
  
      String next() {
          while (!st.hasMoreTokens())
              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());
      }
  
  }
  
  