#include <bits/stdc++.h>
const int mod = 998244353, MAXN = 1e5 + 1;
const long long inf = 1e18;
const double eps = 1e-12;
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  int xors[(int)3e5 + 1];
  xors[0] = 0;
  for (int i = 1; i < 3e5 + 1; i++) {
    xors[i] = (xors[i - 1] ^ i);
  }
  while (t--) {
    int a, b;
    cin >> a >> b;
    int ans = 0;
    int curr = 0;
    ans = a;
    curr = xors[a - 1];
    if (curr == b)
      cout << ans;
    else if ((curr ^ b) != a)
      cout << ans + 1;
    else
      cout << ans + 2;
    cout << '\n';
  }
}

### 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>
const int mod = 998244353, MAXN = 1e5 + 1;
const long long inf = 1e18;
const double eps = 1e-12;
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  int xors[(int)3e5 + 1];
  xors[0] = 0;
  for (int i = 1; i < 3e5 + 1; i++) {
    xors[i] = (xors[i - 1] ^ i);
  }
  while (t--) {
    int a, b;
    cin >> a >> b;
    int ans = 0;
    int curr = 0;
    ans = a;
    curr = xors[a - 1];
    if (curr == b)
      cout << ans;
    else if ((curr ^ b) != a)
      cout << ans + 1;
    else
      cout << ans + 2;
    cout << '\n';
  }
}

```