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.
#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long d = 0, w = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') d = d * 10 + ch - '0', ch = getchar();
  return d * w;
}
void write_(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write_(x / 10);
  putchar(x % 10 + '0');
}
void writeln(long long x) {
  write_(x);
  putchar('\n');
}
void writesp(long long x) {
  write_(x);
  putchar(' ');
}
signed main() {
  freopen("data.in", "r", stdin);
  freopen("data.out", "w", stdout);
  long long T = read();
  while (T--) {
    long long n = read(), m = read();
    long long ans = 0;
    if ((n | m) == m) {
      ans = n ^ (m + 1);
    } else {
      for (long long i = 50; i >= 0; --i) {
        long long val = 1ll << i;
        if (!(n & val) && (m & val)) {
          ans ^= val;
        } else if ((n & val) && !(m & val))
          break;
      }
    }
    writeln(ans);
  }
  return 0;
}
