#include <bits/stdc++.h>
using namespace std;
void dbg() { std::cout << "  #\n"; }
template <typename T, typename... Args>
void dbg(T a, Args... args) {
  std::cout << a << ' ';
  dbg(args...);
}
const long long maxn = 2e5 + 10;
const long long MAX = 1000;
const long long inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
long long n, m;
long long a[maxn];
long long b[maxn];
long long getint() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  long long x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x;
}
void solve() {
  n = getint();
  m = getint();
  long long p = m + 1;
  long long ans = 0;
  for (long long i = (30); i >= (0); --i) {
    long long tmp = 1ll << i;
    if (((n & tmp) && (p & tmp)) || (!(n & tmp) && !(p & tmp))) {
      continue;
    } else if (!(n & tmp) && (p & tmp)) {
      ans |= tmp;
    } else {
      break;
    }
  }
  cout << (ans) << endl;
}
signed main() {
  long long TestCase = 1;
  cin >> TestCase;
  while (TestCase--) {
    solve();
  }
  char EndFile = getchar();
  EndFile = getchar();
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

\operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 

Input

The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.

The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).

Output

For each test case, print a single integer — the answer to the problem.

Example

Input


5
3 5
4 6
3 2
69 696
123456 654321


Output


4
3
0
640
530866

Note

In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.

In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.

In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void dbg() { std::cout << "  #\n"; }
template <typename T, typename... Args>
void dbg(T a, Args... args) {
  std::cout << a << ' ';
  dbg(args...);
}
const long long maxn = 2e5 + 10;
const long long MAX = 1000;
const long long inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
long long n, m;
long long a[maxn];
long long b[maxn];
long long getint() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  long long x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x;
}
void solve() {
  n = getint();
  m = getint();
  long long p = m + 1;
  long long ans = 0;
  for (long long i = (30); i >= (0); --i) {
    long long tmp = 1ll << i;
    if (((n & tmp) && (p & tmp)) || (!(n & tmp) && !(p & tmp))) {
      continue;
    } else if (!(n & tmp) && (p & tmp)) {
      ans |= tmp;
    } else {
      break;
    }
  }
  cout << (ans) << endl;
}
signed main() {
  long long TestCase = 1;
  cin >> TestCase;
  while (TestCase--) {
    solve();
  }
  char EndFile = getchar();
  EndFile = getchar();
  return 0;
}

```