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.math.BigInteger;
  import java.util.*;
  
  public class Main {
      static int MOD = 1000000007;
  
      // After writing solution, quick scan for:
      //   array out of bounds
      //   special cases e.g. n=1?
      //   npe, particularly in maps
      //
      // Big numbers arithmetic bugs:
      //   int overflow
      //   sorting, or taking max, after MOD
      void solve() throws IOException {
          // dp[i][b] is the number of bits b for numbers up to i.
          int[][] dp = new int[200001][18];
          for (int i = 1; i <= 200000; i++) {
              for (int b = 0; b < 18; b++) {
                  dp[i][b] = dp[i-1][b];
                  if (((1 << b) & i) > 0) dp[i][b]++;
              }
          }
  
          int T = ri();
          for (int Ti = 0; Ti < T; Ti++) {
              int[] lr = ril(2);
              int l = lr[0];
              int r = lr[1];
  
              int best = Integer.MAX_VALUE;
              for (int b = 0; b < 18; b++) {
                  // Make bit b all ones.
                  int countOnes = dp[r][b] - dp[l-1][b];
                  best = Math.min(best, r-l+1 - countOnes);
              }
  
              pw.println(best);
          }
      }
      // IMPORTANT
      // DID YOU CHECK THE COMMON MISTAKES ABOVE?
  
      // Template code below
  
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter pw = new PrintWriter(System.out);
  
      public static void main(String[] args) throws IOException {
          Main m = new Main();
          m.solve();
          m.close();
      }
  
      void close() throws IOException {
          pw.flush();
          pw.close();
          br.close();
      }
  
      int ri() throws IOException {
          return Integer.parseInt(br.readLine().trim());
      }
  
      long rl() throws IOException {
          return Long.parseLong(br.readLine().trim());
      }
  
      int[] ril(int n) throws IOException {
          int[] nums = new int[n];
          int c = 0;
          for (int i = 0; i < n; i++) {
              int sign = 1;
              c = br.read();
              int x = 0;
              if (c == '-') {
                  sign = -1;
                  c = br.read();
              }
              while (c >= '0' && c <= '9') {
                  x = x * 10 + c - '0';
                  c = br.read();
              }
              nums[i] = x * sign;
          }
          while (c != '\n' && c != -1) c = br.read();
          return nums;
      }
  
      long[] rll(int n) throws IOException {
          long[] nums = new long[n];
          int c = 0;
          for (int i = 0; i < n; i++) {
              int sign = 1;
              c = br.read();
              long x = 0;
              if (c == '-') {
                  sign = -1;
                  c = br.read();
              }
              while (c >= '0' && c <= '9') {
                  x = x * 10 + c - '0';
                  c = br.read();
              }
              nums[i] = x * sign;
          }
          while (c != '\n' && c != -1) c = br.read();
          return nums;
      }
  
      int[] rkil() throws IOException {
          int sign = 1;
          int c = br.read();
          int x = 0;
          if (c == '-') {
              sign = -1;
              c = br.read();
          }
          while (c >= '0' && c <= '9') {
              x = x * 10 + c - '0';
              c = br.read();
          }
          return ril(x);
      }
  
      long[] rkll() throws IOException {
          int sign = 1;
          int c = br.read();
          int x = 0;
          if (c == '-') {
              sign = -1;
              c = br.read();
          }
          while (c >= '0' && c <= '9') {
              x = x * 10 + c - '0';
              c = br.read();
          }
          return rll(x);
      }
  
      char[] rs() throws IOException {
          return br.readLine().toCharArray();
      }
  
      void sort(int[] A) {
          Random r = new Random();
          for (int i = A.length-1; i > 0; i--) {
              int j = r.nextInt(i+1);
              int temp = A[i];
              A[i] = A[j];
              A[j] = temp;
          }
          Arrays.sort(A);
      }
  
      void printDouble(double d) {
          pw.printf("%.16f", d);
      }
  }