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 <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <sstream>
#include <cstring>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <string.h>
#include <assert.h>
#include <iomanip>
#include <bitset>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ordered_set tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
#define int long long
#define ull unsigned long long
#define F first
#define S second
#define nl "\n"
#define ld long double
#define sortv(v) sort(v.begin(),v.end())
#define rsortv(v) sort(v.rbegin(),v.rend())
#define all(v) v.begin(),v.end()
#define div 1000000007
#define div2 998244353
#define PI 3.141592653589793
#define inf 1000000000000000000LL
#define blk 320
#define pb push_back


void solve()
{
    int n,m;cin>>n>>m;
    if(m<n)
    {
        cout<<0<<nl;
        return;
    }
    int ans=inf;
    int idx=30;
    while(idx>=0 && (m&(1<<idx))==0)
        idx--;
    if(idx<0)
    {
        if(n==0)
            cout<<1<<nl;
        else
            cout<<0<<nl;
        return;
    }
    int cur=0;
    for(int i=idx;i>=0;i--)
    {
        if((m&(1<<i))==0)
        {
            int mex=cur;
            if((n&(1<<i))==0)
                mex+=(1<<i);
            ans=min(ans,mex);
        }
        else
        {
            if((n&(1<<i))==0)
                cur+=(1<<i);
        }
    }
    cout<<ans<<nl;
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif
        int t=1;
        cin>>t;
        while(t--)
        {
            solve();
        }
}