#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e17;
const long long NINF = (-1) * INF;
vector<string> s1;
long long power(long long x, long long y, long long m) {
  long long res = 1;
  x %= m;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y >>= 1;
    x = (x * x) % m;
  }
  return res % m;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
bool isvalid(long long ar, long long n) {
  if (ar > n - 1) return false;
  return true;
}
long long sumOfDigit(long long n, long long b) {
  long long unitDigit, sum = 0;
  while (n > 0) {
    unitDigit = n % b;
    sum += unitDigit;
    n = n / b;
  }
  return sum;
}
long long computeXOR(long long n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
class sol {
 public:
  void solve() {
    long long a, b;
    cin >> a >> b;
    long long n = a;
    if (a > 1)
      a = computeXOR(a - 1);
    else
      a = 0;
    if (a == b) {
      cout << n << endl;
    } else {
      a ^= b;
      if (a == n)
        cout << n + 2 << endl;
      else
        cout << n + 1 << endl;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  sol s;
  while (t--) s.solve();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e17;
const long long NINF = (-1) * INF;
vector<string> s1;
long long power(long long x, long long y, long long m) {
  long long res = 1;
  x %= m;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y >>= 1;
    x = (x * x) % m;
  }
  return res % m;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
bool isvalid(long long ar, long long n) {
  if (ar > n - 1) return false;
  return true;
}
long long sumOfDigit(long long n, long long b) {
  long long unitDigit, sum = 0;
  while (n > 0) {
    unitDigit = n % b;
    sum += unitDigit;
    n = n / b;
  }
  return sum;
}
long long computeXOR(long long n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
class sol {
 public:
  void solve() {
    long long a, b;
    cin >> a >> b;
    long long n = a;
    if (a > 1)
      a = computeXOR(a - 1);
    else
      a = 0;
    if (a == b) {
      cout << n << endl;
    } else {
      a ^= b;
      if (a == n)
        cout << n + 2 << endl;
      else
        cout << n + 1 << endl;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  sol s;
  while (t--) s.solve();
  return 0;
}

```