#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
long long int dp[2000002];
long long int po[2000002];
long long int gcdExtended(long long int a, long long int b, long long int* x,
                          long long int* y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcdExtended(a, m, &x, &y);
  long long int res = (x % m + m) % m;
  return res;
}
void init(string s) {
  long long int n = s.length();
  long long int g = 31;
  po[1] = 1;
  for (long long int i = 1; i <= n; i++) {
    long long int ind = s[i - 1] - 'a' + 1;
    dp[i] = (dp[i - 1] + ind * po[i]) % mod;
    po[i + 1] = (po[i] * g) % mod;
  }
}
long long int get(long long int l, long long int r) {
  long long int ans = (dp[l + r] - dp[l - 1] + mod) % mod;
  long long int iv = modInverse(po[l], mod);
  ans = (ans * iv) % mod;
  return ans;
}
void solve() {
  long long int n, m;
  cin >> n >> m;
  long long int ans = 0;
  long long int l = 31;
  while (l > 1) {
    if (n > m) {
      break;
    }
    for (long long int i = 1; i < 31; i++) {
      long long int x = n;
      x |= ((1 << i) - 1);
      if (x > m) {
        ans += (1 << (i - 1));
        n |= (1 << (i - 1));
        l = i;
        break;
      }
    }
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
long long int mod = 1e9 + 7;
long long int dp[2000002];
long long int po[2000002];
long long int gcdExtended(long long int a, long long int b, long long int* x,
                          long long int* y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcdExtended(a, m, &x, &y);
  long long int res = (x % m + m) % m;
  return res;
}
void init(string s) {
  long long int n = s.length();
  long long int g = 31;
  po[1] = 1;
  for (long long int i = 1; i <= n; i++) {
    long long int ind = s[i - 1] - 'a' + 1;
    dp[i] = (dp[i - 1] + ind * po[i]) % mod;
    po[i + 1] = (po[i] * g) % mod;
  }
}
long long int get(long long int l, long long int r) {
  long long int ans = (dp[l + r] - dp[l - 1] + mod) % mod;
  long long int iv = modInverse(po[l], mod);
  ans = (ans * iv) % mod;
  return ans;
}
void solve() {
  long long int n, m;
  cin >> n >> m;
  long long int ans = 0;
  long long int l = 31;
  while (l > 1) {
    if (n > m) {
      break;
    }
    for (long long int i = 1; i < 31; i++) {
      long long int x = n;
      x |= ((1 << i) - 1);
      if (x > m) {
        ans += (1 << (i - 1));
        n |= (1 << (i - 1));
        l = i;
        break;
      }
    }
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```