#include <bits/stdc++.h>
const unsigned int MOD = 1000000007;
const long double u = 1e+9;
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    int p = m + 1;
    string n1 = "";
    string m1 = "";
    int z = 1;
    int w = 1 << 30;
    while (z < w) {
      int d = z & n;
      if (d == z) {
        n1 = "1" + n1;
      } else {
        n1 = "0" + n1;
      }
      z <<= 1;
    }
    z = 1;
    while (z < w) {
      int d = z & p;
      if (d == z) {
        m1 = "1" + m1;
      } else {
        m1 = "0" + m1;
      }
      z <<= 1;
    }
    string h = "";
    z = 0;
    for (int i = 0; i < n1.size(); i++) {
      if (z == 1) {
        h += "0";
        continue;
      }
      if (n1[i] == m1[i]) {
        h += "0";
        continue;
      }
      if (n1[i] > m1[i]) {
        h += "0";
        z = 1;
        continue;
      }
      if (n1[i] < m1[i]) {
        h += "1";
      }
    }
    int j = 0;
    reverse(h.begin(), h.end());
    for (int i = 0; i < h.size(); i++) {
      j += (h[i] - '0') * pow(2, i);
    }
    cout << j << endl;
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
const unsigned int MOD = 1000000007;
const long double u = 1e+9;
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    int p = m + 1;
    string n1 = "";
    string m1 = "";
    int z = 1;
    int w = 1 << 30;
    while (z < w) {
      int d = z & n;
      if (d == z) {
        n1 = "1" + n1;
      } else {
        n1 = "0" + n1;
      }
      z <<= 1;
    }
    z = 1;
    while (z < w) {
      int d = z & p;
      if (d == z) {
        m1 = "1" + m1;
      } else {
        m1 = "0" + m1;
      }
      z <<= 1;
    }
    string h = "";
    z = 0;
    for (int i = 0; i < n1.size(); i++) {
      if (z == 1) {
        h += "0";
        continue;
      }
      if (n1[i] == m1[i]) {
        h += "0";
        continue;
      }
      if (n1[i] > m1[i]) {
        h += "0";
        z = 1;
        continue;
      }
      if (n1[i] < m1[i]) {
        h += "1";
      }
    }
    int j = 0;
    reverse(h.begin(), h.end());
    for (int i = 0; i < h.size(); i++) {
      j += (h[i] - '0') * pow(2, i);
    }
    cout << j << endl;
  }
}

```