#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mxN = 3e5 + 5, mod = 1e9 + 7, LOG = 20;
int x[mxN];
void solve() {
  int a, b;
  cin >> a >> b;
  if (x[a - 1] == b)
    cout << a;
  else if ((b ^ x[a - 1]) == a)
    cout << a + 2;
  else
    cout << a + 1;
}
int main() {
  ios ::sync_with_stdio(false), cin.tie(nullptr);
  for (int i = 1; i < mxN; ++i) x[i] = (x[i - 1] ^ i);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
using ll = long long;
const int mxN = 3e5 + 5, mod = 1e9 + 7, LOG = 20;
int x[mxN];
void solve() {
  int a, b;
  cin >> a >> b;
  if (x[a - 1] == b)
    cout << a;
  else if ((b ^ x[a - 1]) == a)
    cout << a + 2;
  else
    cout << a + 1;
}
int main() {
  ios ::sync_with_stdio(false), cin.tie(nullptr);
  for (int i = 1; i < mxN; ++i) x[i] = (x[i - 1] ^ i);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}

```