#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
long long updiv(long long a, long long b) {
  if (a % b == 0)
    return (a / b);
  else
    return (a / b) + 1;
}
long long mymod(long long x, long long m) {
  if (x > 0)
    return x % m;
  else
    return ((x) % (m) + (m)) % (m);
}
long long XOR(long long n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
long long G(long long x) {
  long long a = x % 8;
  if (a == 0 || a == 1) return x;
  if (a == 2 || a == 3) return 2;
  if (a == 4 || a == 5) return x + 2;
  if (a == 6 || a == 7) return 0;
  return 0;
}
long long countSetBits(long long n) {
  if (n == 0)
    return 0;
  else
    return 1 + countSetBits(n & (n - 1));
}
long long spf[10000001];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 10000001; i++) spf[i] = i;
  for (long long i = 4; i < 10000001; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 10000001; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 10000001; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> mp;
long long getFactorization(long long x) {
  vector<long long> ret;
  long long ans = 1;
  mp.clear();
  while (x != 1) {
    mp[spf[x]]++;
    x = x / spf[x];
  }
  for (auto P : mp) {
    if (P.second % 2 == 1) ans *= P.first;
  }
  return ans;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long computeXOR(long long n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
void solve() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a, b;
  cin >> a >> b;
  long long ans = computeXOR(a - 1);
  long long fans;
  if (ans == b) {
    fans = a;
  } else {
    if ((ans ^ b) == a) {
      fans = a + 2;
    } else {
      fans = a + 1;
    }
  }
  cout << fans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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 long INF = 1e18;
long long updiv(long long a, long long b) {
  if (a % b == 0)
    return (a / b);
  else
    return (a / b) + 1;
}
long long mymod(long long x, long long m) {
  if (x > 0)
    return x % m;
  else
    return ((x) % (m) + (m)) % (m);
}
long long XOR(long long n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
long long G(long long x) {
  long long a = x % 8;
  if (a == 0 || a == 1) return x;
  if (a == 2 || a == 3) return 2;
  if (a == 4 || a == 5) return x + 2;
  if (a == 6 || a == 7) return 0;
  return 0;
}
long long countSetBits(long long n) {
  if (n == 0)
    return 0;
  else
    return 1 + countSetBits(n & (n - 1));
}
long long spf[10000001];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 10000001; i++) spf[i] = i;
  for (long long i = 4; i < 10000001; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 10000001; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 10000001; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> mp;
long long getFactorization(long long x) {
  vector<long long> ret;
  long long ans = 1;
  mp.clear();
  while (x != 1) {
    mp[spf[x]]++;
    x = x / spf[x];
  }
  for (auto P : mp) {
    if (P.second % 2 == 1) ans *= P.first;
  }
  return ans;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long computeXOR(long long n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
void solve() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a, b;
  cin >> a >> b;
  long long ans = computeXOR(a - 1);
  long long fans;
  if (ans == b) {
    fans = a;
  } else {
    if ((ans ^ b) == a) {
      fans = a + 2;
    } else {
      fans = a + 1;
    }
  }
  cout << fans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```