Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.

A number x is said to be a perfect square if there exists an integer y such that x = y2.

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.

The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.

It is guaranteed that at least one element of the array is not a perfect square.

Output

Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.

Examples

Input

2
4 2


Output

2


Input

8
1 2 4 8 16 32 64 576


Output

32

Note

In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.
#include <bits/stdc++.h>
using namespace std;
template <class C>
void mini(C &a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C &a4, C b4) {
  a4 = max(a4, b4);
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
long long cz[4001001];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  if (0) cout << fixed << setprecision(6);
  long long res = -1e7;
  for (long long i = (0); i <= ((long long)(2000) - 1); i++) {
    cz[i * i] = 1;
  }
  long long n;
  cin >> n;
  for (long long i = (0); i <= ((long long)(n)-1); i++) {
    long long pom;
    cin >> pom;
    (pom);
    if (pom < 0 || cz[pom] == 0) {
      maxi(res, pom);
    }
  }
  cout << res << "\n";
}
