#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.14159265359;
inline long long add(long long a, long long b, long long m) {
  if ((a + b) >= m) return (a + b) % m;
  return a + b;
}
inline long long mul(long long a, long long b, long long m) {
  if ((a * b) < m) return a * b;
  return (a * b) % m;
}
long long power(long long x, long long y, long long m) {
  long long res = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y >> 1;
    x = (x * x) % m;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1, n;
  cin >> t;
  vector<int> xo(300005);
  for (int i = 1; i < xo.size(); i++) {
    xo[i] = i ^ xo[i - 1];
  }
  while (t--) {
    int a, b;
    cin >> a >> b;
    int x = xo[a - 1];
    int y = x ^ b;
    if (x == b) {
      cout << a << endl;
    } else if (y != a) {
      cout << a + 1 << endl;
    } else {
      cout << a + 2 << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const long double pi = 3.14159265359;
inline long long add(long long a, long long b, long long m) {
  if ((a + b) >= m) return (a + b) % m;
  return a + b;
}
inline long long mul(long long a, long long b, long long m) {
  if ((a * b) < m) return a * b;
  return (a * b) % m;
}
long long power(long long x, long long y, long long m) {
  long long res = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y >> 1;
    x = (x * x) % m;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1, n;
  cin >> t;
  vector<int> xo(300005);
  for (int i = 1; i < xo.size(); i++) {
    xo[i] = i ^ xo[i - 1];
  }
  while (t--) {
    int a, b;
    cin >> a >> b;
    int x = xo[a - 1];
    int y = x ^ b;
    if (x == b) {
      cout << a << endl;
    } else if (y != a) {
      cout << a + 1 << endl;
    } else {
      cout << a + 2 << endl;
    }
  }
  return 0;
}

```