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.
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<int> adj[200005], adj_rev[200005];
int xMove[] = {0, 1, 0, -1, 1, 1, -1, -1, 2, 2, -2, -2, 1, 1, -1, -1};
int yMove[] = {1, 0, -1, 0, 1, -1, 1, -1, 1, -1, 1, -1, 2, -2, 2, -2};
int main() {
  cin.tie(0), cout.tie(0);
  ios_base::sync_with_stdio(0);
  cout.setf(ios::fixed);
  cout.precision(10);
  int TC = 1;
  int n, k, m, q;
  cin >> TC;
  int x[300005];
  x[0] = 0;
  for (int i = 1; i <= 300000; i++) x[i] = (i ^ x[i - 1]);
  for (int t1 = 1; t1 <= TC; t1++) {
    int a, b;
    cin >> a >> b;
    int cur_xor = x[a - 1];
    int needed = cur_xor ^ b;
    if (cur_xor == b)
      cout << a << "\n";
    else if (needed != a)
      cout << a + 1 << "\n";
    else {
      cout << a + 2 << "\n";
    }
  }
  return 0;
}
