#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
const long long mod = 1000 * 1000 * 1000 + 7;
const long long inf = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long long mod2 = 998244353;
const long long N = 1000 * 1000 + 10;
const long double pi = 3.141592653589793;
long long power(long long x, long long y, long long p = LLONG_MAX) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long ncr(long long n, long long r, long long m) {
  if (r > n) return 0;
  long long a = 1, b = 1, i;
  for (i = 0; i < r; i++) {
    a = (a * n) % m;
    --n;
  }
  while (r) {
    b = (b * r) % m;
    --r;
  }
  return (a * power(b, m - 2, m)) % m;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  bitset<64> a(n), b(0), c(m + 1);
  for (long long i = 63; i >= 0; i--) {
    if (a[i] == c[i]) continue;
    if (a[i] == 1 && c[i] == 0) {
      break;
    } else {
      b[i] = 1;
    }
  }
  long long ans = 0;
  for (long long i = 63; i >= 0; i--) ans += b[i] * pow(2, i);
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
const long long mod = 1000 * 1000 * 1000 + 7;
const long long inf = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long long mod2 = 998244353;
const long long N = 1000 * 1000 + 10;
const long double pi = 3.141592653589793;
long long power(long long x, long long y, long long p = LLONG_MAX) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long ncr(long long n, long long r, long long m) {
  if (r > n) return 0;
  long long a = 1, b = 1, i;
  for (i = 0; i < r; i++) {
    a = (a * n) % m;
    --n;
  }
  while (r) {
    b = (b * r) % m;
    --r;
  }
  return (a * power(b, m - 2, m)) % m;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  bitset<64> a(n), b(0), c(m + 1);
  for (long long i = 63; i >= 0; i--) {
    if (a[i] == c[i]) continue;
    if (a[i] == 1 && c[i] == 0) {
      break;
    } else {
      b[i] = 1;
    }
  }
  long long ans = 0;
  for (long long i = 63; i >= 0; i--) ans += b[i] * pow(2, i);
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```