Alice gave Bob two integers a and b (a > 0 and b ≥ 0). Being a curious boy, Bob wrote down an array of non-negative integers with \operatorname{MEX} value of all elements equal to a and \operatorname{XOR} value of all elements equal to b.

What is the shortest possible length of the array Bob wrote?

Recall that the \operatorname{MEX} (Minimum EXcluded) of an array is the minimum non-negative integer that does not belong to the array and the \operatorname{XOR} of an array is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all the elements of the array.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 5 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The only line of each test case contains two integers a and b (1 ≤ a ≤ 3 ⋅ 10^5; 0 ≤ b ≤ 3 ⋅ 10^5) — the \operatorname{MEX} and \operatorname{XOR} of the array, respectively.

Output

For each test case, output one (positive) integer — the length of the shortest array with \operatorname{MEX} a and \operatorname{XOR} b. We can show that such an array always exists.

Example

Input


5
1 1
2 1
2 0
1 10000
2 10000


Output


3
2
3
2
3

Note

In the first test case, one of the shortest arrays with \operatorname{MEX} 1 and \operatorname{XOR} 1 is [0, 2020, 2021].

In the second test case, one of the shortest arrays with \operatorname{MEX} 2 and \operatorname{XOR} 1 is [0, 1].

It can be shown that these arrays are the shortest arrays possible.
import java.util.*;
import java.lang.*;
import java.io.*;

//author: ankit_bit

public class Main {


    static FastReader fr = new FastReader();
    static StringBuilder sb = new StringBuilder();

    public static void main(String[] args) throws java.lang.Exception {

        int testcases = fr.nextInt();
        long arr[]= new long[300001];
        arr[0]=0;
        for(int i=1;i<300001;i++){
            arr[i]=arr[i-1]^i;
        }
        while (testcases-- > 0) {

            solve(arr);
        }
        System.out.println(sb);

    }

    public static void solve(long arr[]) {
      int mex= fr.nextInt();
      int xor= fr.nextInt();
      long cnt=0, currxor=0;
//      for(int i=0;i<mex;i++){
//          cnt++;
//          currxor^=i;
//      }
      cnt=mex;
      currxor=arr[(int) (cnt-1)];
      if(currxor==xor){
          sb.append(cnt);
      }
      else{
          if((currxor^xor)==mex)
          sb.append(cnt+2);
          else{
              sb.append(cnt+1);
          }
      }
      sb.append("\n");
    }

    static boolean isPrime(int x) {
        for (int d = 2; d * d <= x; d++) {
            if (x % d == 0)
                return false;
        }
        return true;
    }

    static int lowerBound(int a[], int low, int high, int x) {
        int pos = -1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (a[mid] == x) {
                pos = mid;
                high = mid - 1;
            } else if (a[mid] < x) low = mid + 1;
            else high = mid - 1;
        }
        if (pos != -1) return pos;
        return high + 1;
    }

    static int upperBound(int a[], int low, int high, int x) {
        int pos = -1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (a[mid] == x) {
                pos = mid;
                low = mid + 1;
            } else if (a[mid] < x) low = mid + 1;
            else high = mid - 1;
        }
        if (pos != -1) return pos;
        return high + 1;
    }

    static long mult(long a, long b, long mod) {
        long res = (a * b) % mod;
        if (res < 0) {
            res += mod;
        }
        return res;
    }

    static long add(long a, long b, long mod) {
        long res = (a + b) % mod;
        if (res < 0) {
            res += mod;
        }
        return res;
    }

    static long gcd(long a, long b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    static long fact[] = new long[100001], M = 1000000007;

    static long ncr(int n, int r) {
        if (r > n) return 0;
        return fact[n] * power(fact[r], M - 2) % M * power(fact[n - r], M - 2) % M;
    }

    static long power(long a, long b) {
        long res = 1;
        while (b != 0) {
            if (b % 2 == 1) res = res * a % M;
            b >>= 1;
            a = a * a % M;
        }
        return res;
    }


    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i = 0; i < a.length; i++) {
            l.add(a[i]);
        }
        Collections.sort(l);
        for (int i = 0; i < l.size(); i++)
            a[i] = l.get(i);
    }

}

class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
        br = new BufferedReader(new
                InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            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());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}
