#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 15;
const int mod = 1e9 + 7;
inline int read() {
  int ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
inline long long readll() {
  long long ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline int Max(const int e, const int f) { return e > f ? e : f; }
int xr[N];
int main() {
  for (register int i = 1; i <= 3e5; i++) xr[i] = xr[i - 1] ^ i;
  int t = read();
  while (t--) {
    int a = read();
    int b = read();
    if (xr[a - 1] == b)
      printf("%d\n", a);
    else {
      if ((xr[a - 1] ^ b) == a) {
        printf("%d\n", a + 2);
      } else
        printf("%d\n", a + 1);
    }
  }
  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 int N = 3e5 + 15;
const int mod = 1e9 + 7;
inline int read() {
  int ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
inline long long readll() {
  long long ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline int Max(const int e, const int f) { return e > f ? e : f; }
int xr[N];
int main() {
  for (register int i = 1; i <= 3e5; i++) xr[i] = xr[i - 1] ^ i;
  int t = read();
  while (t--) {
    int a = read();
    int b = read();
    if (xr[a - 1] == b)
      printf("%d\n", a);
    else {
      if ((xr[a - 1] ^ b) == a) {
        printf("%d\n", a + 2);
      } else
        printf("%d\n", a + 1);
    }
  }
  return 0;
}

```