/*
 *   BISMILLAHIR RAHMANIR RAHIM
 *   ==========================
 *
 *   Submitted By: SAKLAN
 *   North East University Bangladesh



 */

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;

#define int long long
#define ll long long
#define ld long double
#define pb(a) push_back(a);
#define cinv(v) for(auto &i:v) cin >> i;
#define vi vector<int>
#define vii vector<ll>
#define modd 1000000007
#define coutv(v) for(auto e:v) cout << e << ' ';
#define srt(v) sort(v.begin(),v.end())
#define rsrt(v) sort(v.rbegin(),v.rend())
#define yes cout<<"Yes\n"
#define YES cout<<"YES\n"
#define NO cout<<"NO\n"
#define no cout<<"No\n"
#define mem(a,b) memset(a, b, sizeof(a) )
#define sqr(a) ((a) * (a))
#define comp [](const pair<int,int> &a, const pair<int,int> &b){ return a.second > b.second; }
#define file() freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#define endl '\n'
#define all(v) (v).begin(),(v).end()
#define saklan ios::sync_with_stdio(0); cin.tie(0);
#define dbg(v) do { \
    for (auto val : v) cout << val << " "; \
    cout << endl << "___________________________ " << endl; \
} while(0)

#define otp(v) do { \
    for (auto val : v) cout << val << " "; \
    cout << endl; \
} while(0)

ll gcd ( ll a, ll b ) { return __gcd ( a, b ); }
ll lcm ( ll a, ll b ) { return a * ( b / gcd ( a, b ) ); }



//template<class T>
//using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_nodde_update>;

bool cmp(const pair<ll,ll> &a, const pair<ll,ll> &b) {
    return (a.first - a.second) > (b.first - b.second);
}

bool cmpp(const pair<int,int>& a, const pair<int,int>& b){
    if (a.first != b.first) return a.first < b.first;
    return a.second > b.second;
}

int nCr(int n, int r) {
    if (r > n) return 0;
    if (r > n - r) r = n - r;

    int res = 1;
    for (int i = 0; i < r; i++) {
        res = res * (n - i) / (i + 1);
    }
    return res;
}





void solve(){
//    int n;cin >> n;
//    string s,p;cin >> s >> p;
//    if(s==p){
//        cout << "YES" << endl;
//        return;
//    }
////    int cn = count(p.begin(),p.end(),'1');
////    if(cn)YES;
////    else NO;
//    for(int i = 0; i < n-1; i++){
//        if(s[i] == p[i]){
//            continue;
//        }else{
//            if(s[i] == '0'){
//                s[i] = '1';
//                s[i+1] = '0';
//            }else{
//                s[i] = '0';
//                s[i+1] = '1';
//            }
//        }
//    }
//    if(s==p) YES;
////    if(s[n-1] == p[n-1]) YES;
////    else if(s==p) YES;
//    else NO;

    int n;cin >>n;
    // quadretic sequentional edquation simplify korlam. e problem e case D = 9 + 4 *2k ;

    int d = 9 + (4 * (2*n));
    //cout << d << endl;

    int sq = sqrt(d);
    if(sq*sq==d) cout << "b" << endl;
    else cout <<"a"<<endl;
}



int32_t main() {
    saklan
    #ifndef ONLINE_JUDGE
        //freopen("input.txt", "r", stdin);
    #endif
    int t = 1;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}