#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
void f() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
queue<pair<int, int>> q;
int n, m, ai, aj;
bool vis[2005][2005] = {};
bool lev[2005][2005];
int h[] = {1, 0, 0, -1};
int v[] = {0, -1, 1, 0};
bool check(int i, int j) {
  if (i > 0 && i <= n && j > 0 && j <= m) {
    return 1;
  }
  return 0;
}
void dfs(int i, int j) {
  vis[i][j] = 1;
  for (int ii = 0; ii < 9; ii++) {
  }
}
void bfs() {
  while (!q.empty()) {
    int f = q.front().first, s = q.front().second;
    q.pop();
    vis[f][s] = 1;
    for (int ii = 0; ii < 4; ii++) {
      cout << check(f + h[ii], s + v[ii]);
      if (check(f + h[ii], s + v[ii]) && vis[f + h[ii]][s + v[ii]] == 0) {
        vis[f + h[ii]][s + v[ii]] = 1;
        q.push({f + h[ii], s + v[ii]});
        lev[f + h[ii]][s + v[ii]] = lev[f][s] + 1;
        ai = f + h[ii];
        aj = s + v[ii];
        cout << ai << " " << aj << endl;
      }
    }
  }
}
int main() {
  f();
  long long arr[300005] = {};
  for (int i = 1; i <= 3e5; i++) {
    arr[i] = arr[i - 1] ^ i;
  }
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    if (arr[n] == m)
      cout << n + 2;
    else if (arr[n - 1] == m) {
      cout << n;
    } else {
      cout << n + 1;
    }
    cout << endl;
  }
}

### Prompt

Please formulate 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;
const long long mod = 1e9 + 7;
void f() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
queue<pair<int, int>> q;
int n, m, ai, aj;
bool vis[2005][2005] = {};
bool lev[2005][2005];
int h[] = {1, 0, 0, -1};
int v[] = {0, -1, 1, 0};
bool check(int i, int j) {
  if (i > 0 && i <= n && j > 0 && j <= m) {
    return 1;
  }
  return 0;
}
void dfs(int i, int j) {
  vis[i][j] = 1;
  for (int ii = 0; ii < 9; ii++) {
  }
}
void bfs() {
  while (!q.empty()) {
    int f = q.front().first, s = q.front().second;
    q.pop();
    vis[f][s] = 1;
    for (int ii = 0; ii < 4; ii++) {
      cout << check(f + h[ii], s + v[ii]);
      if (check(f + h[ii], s + v[ii]) && vis[f + h[ii]][s + v[ii]] == 0) {
        vis[f + h[ii]][s + v[ii]] = 1;
        q.push({f + h[ii], s + v[ii]});
        lev[f + h[ii]][s + v[ii]] = lev[f][s] + 1;
        ai = f + h[ii];
        aj = s + v[ii];
        cout << ai << " " << aj << endl;
      }
    }
  }
}
int main() {
  f();
  long long arr[300005] = {};
  for (int i = 1; i <= 3e5; i++) {
    arr[i] = arr[i - 1] ^ i;
  }
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    if (arr[n] == m)
      cout << n + 2;
    else if (arr[n - 1] == m) {
      cout << n;
    } else {
      cout << n + 1;
    }
    cout << endl;
  }
}

```