#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y);
long long int gcd(long long int a, long long int b);
vector<long long int> ft = {1};
long long int modInv(long long int i) { return power(i, 1000000007 - 2); }
long long int ncr(long long int n, long long int r) {
  return (n >= r ? (ft[n] * modInv(ft[r])) % 1000000007 * modInv(ft[n - r]) %
                       1000000007
                 : 0);
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x %= 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int xx[300005];
  xx[0] = 0;
  for (long long int i = 1; i < 300005; i++) {
    xx[i] = xx[i - 1] ^ i;
  }
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    long long int x = xx[a - 1];
    long long int req = b ^ x;
    if (req == 0) {
      cout << a << "\n";
    } else if (req == a) {
      cout << a + 2 << "\n";
    } else {
      cout << a + 1 << "\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>
using namespace std;
long long int power(long long int x, long long int y);
long long int gcd(long long int a, long long int b);
vector<long long int> ft = {1};
long long int modInv(long long int i) { return power(i, 1000000007 - 2); }
long long int ncr(long long int n, long long int r) {
  return (n >= r ? (ft[n] * modInv(ft[r])) % 1000000007 * modInv(ft[n - r]) %
                       1000000007
                 : 0);
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x %= 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int xx[300005];
  xx[0] = 0;
  for (long long int i = 1; i < 300005; i++) {
    xx[i] = xx[i - 1] ^ i;
  }
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    long long int x = xx[a - 1];
    long long int req = b ^ x;
    if (req == 0) {
      cout << a << "\n";
    } else if (req == a) {
      cout << a + 2 << "\n";
    } else {
      cout << a + 1 << "\n";
    }
  }
}

```