Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long double sqr(long double x) { return x * x; }
inline int rand1(int P) {
  long double sum = 0;
  long double x = (long double)rand() / RAND_MAX;
  long double frac = 1;
  for (int k = 0; k <= 2 * P; ++k) {
    if (k) {
      frac *= k;
    }
    sum += (pow(P, k) * exp(-P)) / frac;
    if (x <= sum) {
      return k;
    }
  }
  return 2 * P;
}
inline int rand2(int P) {
  long double x = (long double)rand() / RAND_MAX;
  return (int)floor(x * (2 * P + 1));
}
int a[250];
int _cnt[500], *cnt = _cnt + 100;
long double miss1(int n) {
  if (a[0] < -n) {
    return 1e100;
  }
  long double frac = 1, sum = 0;
  for (int k = 0; k < 100; ++k) {
    if (k) {
      frac *= k;
    }
    sum += sqr((pow(n, k) * exp(-n)) / frac * 250 - cnt[k - n]);
  }
  return sum;
}
long double miss2(int n) {
  if (a[0] < -n || a[249] > n) {
    return 1e100;
  }
  long double sum = 0;
  for (int k = -n; k <= n; ++k) {
    sum += sqr(cnt[k] - 250.0 / (n * 2 + 1));
  }
  return sum;
}
int main(void) {
  int cas;
  scanf("%d", &cas);
  while (cas--) {
    for (int i = 0; i < 250; ++i) {
      scanf("%d", a + i);
    }
    sort(a, a + 250);
    if (a[0] <= -20 || a[249] >= 300) {
      long double fangcha = 0;
      for (int i = 0; i < 250; ++i) {
        fangcha += sqr(a[i]);
      }
      fangcha /= 250;
      if (fangcha < sqr(max(a[249], -a[0])) / 4) {
        puts("poisson");
      } else {
        puts("uniform");
      }
      continue;
    }
    memset(_cnt, 0, sizeof _cnt);
    for (int i = 0; i < 250; ++i) {
      ++cnt[a[i]];
    }
    bool ans;
    long double best = 1e100;
    for (int n = 1; n <= 50; ++n) {
      long double tmp = miss1(n);
      if (tmp < best) {
        best = tmp;
        ans = true;
      }
    }
    for (int n = 1; n <= 50; ++n) {
      long double tmp = miss2(n);
      if (tmp < best) {
        best = tmp;
        ans = false;
      }
    }
    puts(ans ? "poisson" : "uniform");
  }
  return 0;
}