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;
template <typename... Args>
void logger(string vars, Args&&... values) {}
template <typename T>
inline void maxa(T& a, T b) {
  a = max(a, b);
}
template <typename T>
inline void mina(T& a, T b) {
  a = min(a, b);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int mxN = 3e5 + 5;
const int mod = 1e9 + 7;
int pref[mxN];
void solve() {
  int a, b;
  cin >> a >> b;
  int res = pref[a - 1];
  if ((res ^ b) == 0) {
    cout << a << "\n";
  } else if ((res ^ b) == a) {
    cout << a + 2 << "\n";
  } else {
    cout << a + 1 << "\n";
  }
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 1; i <= 3e5; i++) {
    pref[i] = pref[i - 1] ^ i;
  }
  cin >> t;
  while (t--) solve();
}
