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>
using namespace std;
//debug
#ifdef grief
#define debug(...) do{\
	fprintf(stderr , "%s - %d : (%s) = " , __PRETTY_FUNCTION__ , __LINE__ , #__VA_ARGS__ );\
	_DO(__VA_ARGS__);\
}while(0)
template<typename I> 				void _DO(I&&x){	cerr<<x<<endl;	}
template<typename I,typename...T>	void _DO(I&&x,T&&...tail){	cerr<<x<<" , ";	_DO(tail...);	}
template<typename _a,typename _b> 	ostream& operator << (ostream &_s,const pair<_a,_b> &_p)
{	return _s<<'('<<_p.first<<','<<_p.second<<')';	}
#define IOS
#else
#define debug(...)
#define IOS do{ios_base::sync_with_stdio(0); cin.tie(0);}while(0);
#endif
//type
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef priority_queue<pll,vector<pll>,greater<pll> > heap;
//macro
#define SZ(x) ((ll)(x).size())
#define ALL(x) (x).begin(),(x).end()
#define F first
#define S second
#define pb push_back
#define eb emplace_back
#define stp setprecision(30)<<fixed
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll MAX=1e5+5,Mlg=__lg(MAX)+2;
const ll MOD=1e9+7;
//}}}
double arr[260];
int main(){
	IOS
	int n;	cin>>n;
	while(n--){
	    double ma = 0;
		double sum=0;
		for(int i=0;i<250;i++){
			cin>>arr[i];
			sum+=arr[i];
			ma = max(ma, arr[i]);
		}
		double mu=sum/250;
		double var=0;
		for(int i=0;i<250;i++)
			var+=(arr[i])*(arr[i]);
		var/=250;
		if(ma / sqrt(var) > 1.95 )  {
			cout<<"poisson\n";
		}
		else
			cout<<"uniform\n";
	}
	return 0;
}