#include <bits/stdc++.h>
using namespace std;
int num[100100];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &num[i]);
  sort(num, num + n);
  for (int i = n - 1; i >= 0; i--) {
    if (num[i] >= 0) {
      int a = sqrt(num[i]);
      if (a * a != num[i]) {
        printf("%d\n", num[i]);
        break;
      }
    } else {
      printf("%d\n", num[i]);
      break;
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int num[100100];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &num[i]);
  sort(num, num + n);
  for (int i = n - 1; i >= 0; i--) {
    if (num[i] >= 0) {
      int a = sqrt(num[i]);
      if (a * a != num[i]) {
        printf("%d\n", num[i]);
        break;
      }
    } else {
      printf("%d\n", num[i]);
      break;
    }
  }
  return 0;
}

```