description: You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?
  
  A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
  
  The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.
  
  Output
  
  For each test case, output a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  1 2
  2 8
  4 5
  1 5
  100000 200000
  
  
  Output
  
  
  1
  3
  0
  2
  31072
  
  Note
  
  In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.
  
  In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.
solution: import java.io.*;
  import java.util.*;
  
  public class B {
      static int mod = (int) 1e9 + 7;
  
      public static void main(String[] args) throws IOException {
  
          int[][] dp = new int[200005][20];
          for (int i = 1; i <= 200000; i++) {
              int temp = i;
              int index = 0;
              while (temp > 0) {
                  dp[i][index] += (temp % 2);
                  index++;
                  temp /= 2;
              }
  
              for (int j = 0; j < 20; j++)
                  dp[i][j] += dp[i - 1][j];
  
          }
  
          BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
          int T = Integer.parseInt(bufferedReader.readLine());
          while (T-- > 0) {
  
              StringTokenizer stringTokenizer = new StringTokenizer(bufferedReader.readLine());
              int a = Integer.parseInt(stringTokenizer.nextToken());
              int b = Integer.parseInt(stringTokenizer.nextToken());
  
              long ans=0;
              for (int i = 0; i < 20; i++) {
                  ans = Math.max(dp[b][i]-dp[a-1][i], ans);
              }
  
              out.println(b-a+1-ans);
  
          }
          out.close();
      }
  
      public static long power(long a, int p) {
          long res = 1;
          while (p > 0) {
              if ((p & 1) == 1)
                  res = (res * a) % mod;
              a = (a * a) % mod;
              p >>= 1;
          }
          return res;
      }
  
      static long gcd(long a, long b) {
          if (a == 0)
              return b;
          return gcd(b % a, a);
      }
  
      static long lcm(int a, int b) {
          return (a / gcd(a, b)) * b;
      }
  }
  
  /*
   * boolean[] sieve = new boolean[1_000_001];
   * Arrays.fill(sieve, true);
   * sieve[0] = false;
   * sieve[1] = false;
   * 
   * for (int x = 2; x * x < sieve.length; x++) {
   * if (sieve[x]) {
   * for (int j = x; j * x < sieve.length; j++) {
   * sieve[j * x] = false;
   * }
   * }
   * }
   */
  