Time is limited.

//A
#include<iostream>

using namespace std;

int main()
{
    long long n,k;
    while(cin>>n>>k){
        if((n&k)==k){
            cout<<"1"<<endl;
        }else{
            cout<<"0"<<endl;
        }
    }
    return 0;
}
//B
#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N=1e9;
const int mod=998244353;
ll fac[1010];

void init(){
    fac[0]=1;
    for(int i=1;i<=1010;i++){
        fac[i]=fac[i-1]*i%mod;
    }
}
ll ans(int n){
    if(n&1){
        return 0;
    }else{
        return (fac[n/2]*fac[n/2])%mod;
    }
}
int main(){
    int t;
    cin>>t;
    init();
    while(t--){
        int n;
        cin>>n;
        cout<<ans(n)<<endl;
    }
    return 0;
}
//E 题你写过的原题