Your task is the exact same as for the easy version. But this time, the marmots subtract the village's population P from their random number before responding to Heidi's request.

Also, there are now villages with as few as a single inhabitant, meaning that <image>.

Can you help Heidi find out whether a village follows a Poisson or a uniform distribution?

Input

Same as for the easy and medium versions. But remember that now 1 ≤ P ≤ 1000 and that the marmots may provide positive as well as negative integers.

Output

Output one line per village, in the same order as provided in the input. The village's line shall state poisson if the village's distribution is of the Poisson type, and uniform if the answers came from a uniform distribution.

 #include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(const auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define sar(a,n) {cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl;}
 
using namespace std;
 
template<typename S,typename T>auto&operator<<(ostream&o,pair<S,T>p){return o<<"{"<<p.fi<<","<<p.se<<"}";}
template<typename T>auto&operator<<(ostream&o,set<T>s){for(auto&e:s)o<<e<<" ";return o;}
template<typename S,typename T,typename U>
auto&operator<<(ostream&o,priority_queue<S,T,U>q){while(!q.empty())o<<q.top()<<" ",q.pop();return o;}
template<typename K,typename T>auto&operator<<(ostream&o,map<K,T>&m){for(auto&e:m)o<<e<<" ";return o;}
template<typename T>auto&operator<<(ostream&o,vector<T>v){for(auto&e:v)o<<e<<" ";return o;}
void ashow(){cout<<endl;}template<typename T,typename...A>void ashow(T t,A...a){cout<<t<<" ";ashow(a...);}
template<typename S,typename T,typename U>
struct TRI{S fi;T se;U th;TRI(){}TRI(S f,T s,U t):fi(f),se(s),th(t){}
bool operator<(const TRI&_)const{return(fi==_.fi)?((se==_.se)?(th<_.th):(se<_.se)):(fi<_.fi);}};
template<typename S,typename T,typename U>
auto&operator<<(ostream&o,TRI<S,T,U>&t){return o<<"{"<<t.fi<<","<<t.se<<","<<t.th<<"}";}
 
typedef pair<int, int> P;
typedef pair<ll,ll> pll;
typedef TRI<int, int, int> tri;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;
 
const int MAX_N = 100005;
 
int bue[MAX_N];
double sm[MAX_N], LOG[MAX_N];
 
double solve1(int k, int p){
    return k * LOG[p] - p - sm[k];
}
 
double solve2(int p){
    return -LOG[2*p+1];
}
 
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    rep(i,MAX_N){
        LOG[i] = log(i);
        if(i) sm[i] = sm[i-1] + LOG[i];
    }
    rep(i,n){
        // double mean = 0;
        rep(j,250){
            cin >> bue[j];
        }
        // mean /= 250.0;
        // double hoge = 0;
        // rep(j,250){
        //     hoge += (bue[j] - mean) * (bue[j] - mean);
        // }
        // hoge /= 250.0;
        bool ans = false;
        double res = -1e300;
        srep(p,max(1, -*min_element(bue, bue + 250)),1001){
            double ch1 = 0, ch2 = 0;
            bool dame = false;
            rep(j,250){
                if(bue[j]+p > 2 * p) dame = true;
                ch1 += solve1(bue[j]+p, p), ch2 += solve2(p);
            }
            if(dame) ch2 = -1e301;
            if(ch1 > res) res = ch1, ans = false;
            if(ch2 > res) res = ch2, ans = true;
        }
        if(ans){
            cout << "uniform\n";
        }else{
            cout << "poisson\n";
        }
        // if(abs(hoge - mean) < abs(hoge - mean * mean / 3.0)){
        //     int ans = -1;
        //     double res = -1e300;
        //     srep(p,10,1001){
        //         double ch = 0;
        //         rep(j,250){
        //             ch += solve1(bue[j], p);
        //         }
        //         if(ch > res) res = ch, ans = p;
        //     }
        //     cout<< ans << "\n";
        // }else{
        //     const int cand = (*max_element(bue, bue + 250) + 1) / 2;
        //     cout << cand << "\n";
            // if(cand < 50){
            // }else{
            //     const int ans = round(mean);
            //     cout << ans << "\n";
            // }
            // double res = -1e300;
            // srep(p,10,1001){
            //     double ch = 0;
            //     rep(j,250){
            //         ch += solve2(p);
            //     }
            //     if(ch > res) ch = res, ans = p;
            // }
        // }
    }
    return 0;
}