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//jak za ten kod bedzie choc jeden punkt to sie uraduje, DRUGA PRÓBA AUUUUUUUUUUUUUUUU
//dzień 3 czekania, aż organizator napisze 67 na ogłoszeniach!
//////////////////////////////////////////////////
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define tak cout<<"TAK\n"
#define yes cout<<"YES\n"
#define nie cout<<"NIE\n"
#define no cout<<"NO\n"
using ll = long long;
using ld = long double;
using pll = pair<ll,ll>;
using pdd = pair<ld,ld>;
#define vec vector
#define all(x) (x).begin(), (x).end()
#define sz(x) (x).size()
#define linijki ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mapa map<ll,ll>
#define umapa unordered_map<ll,ll>
#define se set<ll>
#define ff first
#define ss second
#define str string
#define st stack<ll>
const ll alph = 26;
const ll BIN6 = (1LL << 20);
const ll BIN9 = (1LL << 30);
const ll BIN18 = (1LL << 60);
const ll INF5 = 100000;
const ll INF6 = 1000000;
const ll INF9 = 1000000000;
const ll INF18 = 1000000000000000000;
const ld DINF5 = 1e5L;
const ld DINF6 = 1e6L;
const ld DINF9 = 1e9L;
const ll MOD = 1000000007;

void solve(){
    ll n,k;
    cin>>n>>k;
    if(k == 1){
        if(n == 1){
            cout << "A\n";
        }
        else if(n == 2){
            cout << "AP\n";
        }
        else{
            nie;
        }
        return;
    }
    else if (k == 2) {
        if (n > 4){
            nie;
            return;
        }
        else if(n==2){
            cout<<"AA\n";
        }
        else if(n==3){
            cout<<"AAP\n";
        }
        else if(n==4){
            cout<<"AAPP\n";
        }
        return;
    }
    else if (k == 3) {
        if (n > 8){
            nie;
            return;
        }
        else if(n==3){
            cout<<"AAA\n";
        }
        else if(n==4){
            cout<<"AAAP\n";
        }
        else if(n==5){
            cout<<"AAAPA\n";
        }
        else if(n==6){
            cout<<"AAAPAP\n";
        }
        else if(n==7){
            cout<<"AAAPAPP\n";
        }
        else if(n==8){
            cout<<"AAAPAPPP\n";
        }
        return;
    }
    else {
        str wyn="";
        for(ll i=0;i<k;++i){
            wyn+="A";
        }
        str cykl="PAPPAA";
        for(ll i=k; i<n; ++i){
            ll poz = (i-k+6) % 6;
            wyn += cykl[poz];
        }
        cout<<wyn<<'\n';
        return;
    }
}

int main() {
    linijki;

    ll t;
    cin>>t;
    while(t--){
        solve();
    }

    return 0;
}