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<vector>
#include<map>
#include<set>
#include<cassert>
#include<cassert>
#include<unordered_map>
#include<unordered_set>
#include<functional>
#include<queue>
#include<stack>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<sstream>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
#include<numeric>
#include<random>
#include<chrono>
#include<bitset>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define all(x) (x).begin(), (x).end()
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define sz(x) (int)(x).size()
#define gc getchar
#define IO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mp make_pair

#ifndef ONLINE_JUDGE
#  define LOG(x) (cerr << #x << " = " << (x) << endl)
#else
#  define LOG(x) ((void)0)
#endif

using ll = long long;
using ull = unsigned long long ;
using ld = long double ;
using str = string;
using ordered_set=tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update>;

const double PI=acos(-1);
const ll INF = 1LL<<62;
const ll MINF = -(1LL<<62);

template<typename T> T getint() {
	T val=0;
	char c;
	
	bool neg=false;
	while((c=gc()) && !(c>='0' && c<='9')) {
		neg|=c=='-';
	}

	do {
		val=(val*10)+c-'0';
	} while((c=gc()) && (c>='0' && c<='9'));

	return val*(neg?-1:1);
}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); uniform_int_distribution<int>(0, n-1)(rng)


int main() {
	IO;
	int T;
	cin>>T;
	while(T--) {
		int n;
		cin>>n;
		string s;
		if(n<=26 && 0) {
			for(int i=0;i<n;++i) {
				s+=i+'a';
			}
		}else {
			n-=1;
			vector<int> lst={1};
			while(n>0) {
				lst.pb((n>1?(n/2)*2-1:1));
				n-=lst.back();
			}
			
			string odd, even;
			for(int i=0;i<sz(lst);++i) {
				odd+=string(lst[i]/2+1, 'a'+i);
				even+=string(lst[i]/2, 'a'+i);
			}
			
			reverse(all(even));
			s=odd+even;
		}
		cout<<s<<"\n";
	}
	return 0;
}
