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 T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
bool compare(T x, U y) {
  return (abs(x - y) <= 1e-9);
}
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int ans = INT_MIN;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x >= 0) {
      double chk1 = sqrt(x);
      if (chk1 == (int)chk1) {
        continue;
      } else
        ans = max(ans, x);
    } else
      ans = max(ans, x);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}
