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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = 1e-9;
const long long mod = 1000000007;
const int MAXN = 100000;
void solve() {
  ll n, m;
  cin >> n >> m;
  if (n > m)
    cout << 0 << endl;
  else {
    int msb1 = 0, msb2 = 0;
    for (int i = 0, nnnn = (31); i < nnnn; ++i) {
      if (m & (1ll << i)) {
        msb1 = i;
      }
    }
    for (int i = 0, nnnn = (31); i < nnnn; ++i) {
      if (n & (1ll << i)) {
        msb2 = i;
      }
    }
    ll ans1 = 0;
    bool add = true;
    for (int i = msb2 + 1, nnnn = (msb1 + 1); i < nnnn; ++i) {
      if ((1 << i) & (m)) {
        ans1 += (1 << i) & (m);
        if (i != msb1) add = false;
      }
    }
    if (add) {
      ll temp = 0;
      for (int i = msb2; i >= 0; i--) {
        if (!(n & (1 << i))) temp += (1 << i);
      }
      for (int i = 0, nnnn = (msb2 + 1); i < nnnn; ++i) {
        if (temp & m & (1 << i)) ans1 += (1 << i);
      }
    }
    if ((ans1 ^ n) == m) ans1++;
    cout << ans1 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(12) << fixed;
  int t;
  cin >> t;
  while (t--) solve();
}
