description: You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  
  \operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.
  
  The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).
  
  Output
  
  For each test case, print a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  3 5
  4 6
  3 2
  69 696
  123456 654321
  
  
  Output
  
  
  4
  3
  0
  640
  530866
  
  Note
  
  In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.
  
  In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.
  
  In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.
solution: import java.util.Scanner;
  
  public class C {
      public static void main(String[] args) {
          Scanner scn = new Scanner(System.in);
          int test = scn.nextInt();
          while (test-- > 0) {
              int n = scn.nextInt();
              int m = scn.nextInt();
              if (m < n) {
                  System.out.println(0);
                  continue;
              } else {
                  m++;
                  char[] mS = Integer.toBinaryString(m).toCharArray();
                  char[] nS = Integer.toBinaryString(n).toCharArray();
                  int lastInd = 0;
                  for (int i = 1; i <= nS.length; i++) {
                      char left = nS[nS.length - i];
                      char right = mS[mS.length - i];
                      if (left == '1' && right == '0') {
                          lastInd = i;
                      }
  
                  }
                  m >>= lastInd;
                  m <<= lastInd;
                  n >>= lastInd;
                  n <<= lastInd;
  
                  System.out.println(m ^ n);
              }
          }
  
      }
  }
  