#include <bits/stdc++.h>
using namespace std;
using lli = long long int;
using pii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vlli = vector<long long int>;
using vpii = vector<pair<int, int>>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vi A(n);
  for (int i = 0; i < n; i++) cin >> A[i];
  sort((A).begin(), (A).end());
  for (int i = n - 1; i >= 0; i--) {
    double temp;
    if (A[i] < 0) {
      cout << A[i];
      break;
    }
    temp = sqrt(A[i]);
    int p = (int)temp;
    if ((temp - p) > 1e-8) {
      cout << A[i];
      break;
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using lli = long long int;
using pii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vlli = vector<long long int>;
using vpii = vector<pair<int, int>>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vi A(n);
  for (int i = 0; i < n; i++) cin >> A[i];
  sort((A).begin(), (A).end());
  for (int i = n - 1; i >= 0; i--) {
    double temp;
    if (A[i] < 0) {
      cout << A[i];
      break;
    }
    temp = sqrt(A[i]);
    int p = (int)temp;
    if ((temp - p) > 1e-8) {
      cout << A[i];
      break;
    }
  }
  return 0;
}

```