#include <bits/stdc++.h>
using namespace std;
long long int M = 1e9 + 1;
long long int MM = 998244353;
template <typename... T>
void read(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(string delimiter, T &&...args) {
  ((cout << args << delimiter), ...);
}
long long int gcd(long long int a, long long int b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void swap(long long int &x, long long int &y) {
  long long int temp = x;
  x = y;
  y = temp;
}
bool revsort(long long int a, long long int b) { return a > b; }
void google(long long int t) { cout << "Case #" << t << ": "; }
void precision(long long int a) { cout << setprecision(a) << fixed; }
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m)) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m)) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
vector<long long int> v(1000000 + 1);
void init() {
  v[0] = 0;
  for (int i = 1; i < 1000000 + 1; i++) {
    v[i] = v[i - 1] ^ i;
  }
}
void solve() {
  long long int a, b;
  read(a, b);
  long long int ctn = a;
  long long int ans = v[a - 1];
  if (ans == b) {
    cout << ctn << "\n";
    return;
  }
  long long int res = ans ^ a;
  if (res == b) {
    ctn += 2;
  } else {
    ctn++;
  }
  cout << ctn << "\n";
}
int main() {
  init();
  int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 M = 1e9 + 1;
long long int MM = 998244353;
template <typename... T>
void read(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(string delimiter, T &&...args) {
  ((cout << args << delimiter), ...);
}
long long int gcd(long long int a, long long int b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void swap(long long int &x, long long int &y) {
  long long int temp = x;
  x = y;
  y = temp;
}
bool revsort(long long int a, long long int b) { return a > b; }
void google(long long int t) { cout << "Case #" << t << ": "; }
void precision(long long int a) { cout << setprecision(a) << fixed; }
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m)) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m)) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
vector<long long int> v(1000000 + 1);
void init() {
  v[0] = 0;
  for (int i = 1; i < 1000000 + 1; i++) {
    v[i] = v[i - 1] ^ i;
  }
}
void solve() {
  long long int a, b;
  read(a, b);
  long long int ctn = a;
  long long int ans = v[a - 1];
  if (ans == b) {
    cout << ctn << "\n";
    return;
  }
  long long int res = ans ^ a;
  if (res == b) {
    ctn += 2;
  } else {
    ctn++;
  }
  cout << ctn << "\n";
}
int main() {
  init();
  int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
}

```