#include <bits/stdc++.h>
using namespace std;
long long re() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long mod = 1000000007;
const int maxn = 4000005;
const int pr = 3;
const int phi = 998244352;
const int dx[8] = {1, 0, 0, -1, -1, -1, 1, 1},
          dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};
int lowbit(int x) { return x & (-x); }
long long qk(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return ans;
}
int c[1000005];
void solve() {
  int _;
  cin >> _;
  for (int i = 1; i <= 300000; i++) {
    c[i] = c[i - 1] ^ i;
  }
  while (_--) {
    int a, b;
    cin >> a >> b;
    if (c[a - 1] == b) {
      cout << a << '\n';
    } else if ((c[a - 1] ^ b) == a) {
      cout << a + 2 << '\n';
    } else {
      cout << a + 1 << '\n';
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  solve();
  return 0;
}

### Prompt

Generate a CPP solution to 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;
long long re() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long mod = 1000000007;
const int maxn = 4000005;
const int pr = 3;
const int phi = 998244352;
const int dx[8] = {1, 0, 0, -1, -1, -1, 1, 1},
          dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};
int lowbit(int x) { return x & (-x); }
long long qk(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return ans;
}
int c[1000005];
void solve() {
  int _;
  cin >> _;
  for (int i = 1; i <= 300000; i++) {
    c[i] = c[i - 1] ^ i;
  }
  while (_--) {
    int a, b;
    cin >> a >> b;
    if (c[a - 1] == b) {
      cout << a << '\n';
    } else if ((c[a - 1] ^ b) == a) {
      cout << a + 2 << '\n';
    } else {
      cout << a + 1 << '\n';
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  solve();
  return 0;
}

```