#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 2;
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long expo(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
vector<long long> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<long long> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long mod_add(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long mod_mul(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long mod_sub(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long mod_div(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
int fact[N];
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mod_mul(fact[i - 1], i, 1000000007);
  }
}
long long mod(long long x) {
  return ((x % 1000000007 + 1000000007) % 1000000007);
}
long long inv(long long x) { return expo(x, 1000000007 - 2, 1000000007); }
long long divide(long long a, long long b) {
  return mod_mul(a, inv(b), 1000000007);
}
long long nCr(long long n, long long r) {
  return divide(fact[n], mod_mul(fact[r], fact[n - r], 1000000007));
}
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;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    int pXor;
    if (a % 4 == 1) {
      pXor = a - 1;
    } else if (a % 4 == 2) {
      pXor = 1;
    } else if (a % 4 == 3) {
      pXor = a;
    } else {
      pXor = 0;
    }
    if (pXor == b) {
      cout << a << '\n';
    } else if ((pXor ^ b) != a) {
      cout << a + 1 << '\n';
    } else {
      cout << a + 2 << '\n';
    }
  }
}

### 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 int N = 2e5 + 2;
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long expo(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
vector<long long> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<long long> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long mod_add(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long mod_mul(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long mod_sub(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long mod_div(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
int fact[N];
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mod_mul(fact[i - 1], i, 1000000007);
  }
}
long long mod(long long x) {
  return ((x % 1000000007 + 1000000007) % 1000000007);
}
long long inv(long long x) { return expo(x, 1000000007 - 2, 1000000007); }
long long divide(long long a, long long b) {
  return mod_mul(a, inv(b), 1000000007);
}
long long nCr(long long n, long long r) {
  return divide(fact[n], mod_mul(fact[r], fact[n - r], 1000000007));
}
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;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    int pXor;
    if (a % 4 == 1) {
      pXor = a - 1;
    } else if (a % 4 == 2) {
      pXor = 1;
    } else if (a % 4 == 3) {
      pXor = a;
    } else {
      pXor = 0;
    }
    if (pXor == b) {
      cout << a << '\n';
    } else if ((pXor ^ b) != a) {
      cout << a + 1 << '\n';
    } else {
      cout << a + 2 << '\n';
    }
  }
}

```