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.io.PrintWriter;
  import java.util.ArrayDeque;
  import java.util.HashMap;
  import java.util.Queue;
  
  public class Solution {
      public static long find(int n, int k) {
          int mod = 1000000007;
          long res = 0;
          while (k > 0) {
              int t = log(k, 2);
              long temp = (long) power(n, t, mod);
              res = (res + (temp % mod)) % mod;
              k -= Math.pow(2, t);
          }
          return res;
      }
      private static int log(int n, int b) {
          int temp = 1;
          int i;
          for (i = 1; b * temp <= n; i++) {
              temp *= b;
          }
          return i - 1;
      }
      private static long power(int n, int p, int mod) {
          if (n == 0) return 0;
          if (p == 0) return 1;
          long res = power(n, p / 2, mod);
          if (p % 2 == 0) return (res * res) % mod;
          else return (((res * res) % mod) * (n)) % mod;
      }
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter pw = new PrintWriter(System.out);
          int t = Integer.parseInt(br.readLine());
          for (int j = 0; j < t; j++) {
              String[] s = br.readLine().split("\\s");
              int n = Integer.parseInt(s[0]);
              int k = Integer.parseInt(s[1]);
              pw.println(find(n, k));
          }
          pw.flush();
      }
  }