#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 100;
bool isPerfect[maxn];
vector<int> v;
void init() {
  memset(isPerfect, false, sizeof(isPerfect));
  for (int i = 0; i <= 1000; i++) {
    isPerfect[i * i] = true;
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
  }
  sort((v).begin(), (v).end());
  reverse((v).begin(), (v).end());
  for (int i = 0; i <= (int)v.size() - 1; i++) {
    if ((v[i] < 0) or (!isPerfect[v[i]])) {
      printf("%d\n", v[i]);
      return;
    }
  }
}
int main() {
  init();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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;
const int maxn = 1e6 + 100;
bool isPerfect[maxn];
vector<int> v;
void init() {
  memset(isPerfect, false, sizeof(isPerfect));
  for (int i = 0; i <= 1000; i++) {
    isPerfect[i * i] = true;
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
  }
  sort((v).begin(), (v).end());
  reverse((v).begin(), (v).end());
  for (int i = 0; i <= (int)v.size() - 1; i++) {
    if ((v[i] < 0) or (!isPerfect[v[i]])) {
      printf("%d\n", v[i]);
      return;
    }
  }
}
int main() {
  init();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```