#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const double pai = acos(-1.0);
const int maxn = 1e6 + 10;
const int mod = 998244353;
const double eps = 1e-9;
const int N = 5e3 + 10;
inline int read() {
  int k = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) k = (k << 1) + (k << 3) + c - 48, c = getchar();
  return k * f;
}
long long n, m, mul[40];
int t;
int main() {
  mul[0] = 1;
  for (int i = 1; i <= 32; i++) mul[i] = mul[i - 1] * 2;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &n, &m);
    long long sum = 0, p = m + 1;
    for (int i = 32; i >= 0; i--) {
      long long x = n & (1ll << i), y = p & (1ll << i);
      if (x && !y) break;
      if (y && !x) sum += mul[i];
    }
    printf("%lld\n", sum);
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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>
#pragma GCC optimize(2)
using namespace std;
const double pai = acos(-1.0);
const int maxn = 1e6 + 10;
const int mod = 998244353;
const double eps = 1e-9;
const int N = 5e3 + 10;
inline int read() {
  int k = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) k = (k << 1) + (k << 3) + c - 48, c = getchar();
  return k * f;
}
long long n, m, mul[40];
int t;
int main() {
  mul[0] = 1;
  for (int i = 1; i <= 32; i++) mul[i] = mul[i - 1] * 2;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &n, &m);
    long long sum = 0, p = m + 1;
    for (int i = 32; i >= 0; i--) {
      long long x = n & (1ll << i), y = p & (1ll << i);
      if (x && !y) break;
      if (y && !x) sum += mul[i];
    }
    printf("%lld\n", sum);
  }
  return 0;
}

```