#include <bits/stdc++.h>
using namespace std;
bool isprime(long long n) {
  if (n <= 1) {
    return false;
  }
  if (n == 2 || n == 3) {
    return true;
  }
  if (n % 2 == 0 || n % 3 == 0) {
    return false;
  }
  for (long long i = 5; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  unordered_set<long long> second;
  for (long long i = 0; i <= 1000; i++) {
    second.insert((i * i));
  }
  long long n;
  cin >> n;
  long long ans = -1000000000;
  while (n--) {
    long long x;
    cin >> x;
    if (second.count(x) == 0) {
      ans = max(ans, x);
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
bool isprime(long long n) {
  if (n <= 1) {
    return false;
  }
  if (n == 2 || n == 3) {
    return true;
  }
  if (n % 2 == 0 || n % 3 == 0) {
    return false;
  }
  for (long long i = 5; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  unordered_set<long long> second;
  for (long long i = 0; i <= 1000; i++) {
    second.insert((i * i));
  }
  long long n;
  cin >> n;
  long long ans = -1000000000;
  while (n--) {
    long long x;
    cin >> x;
    if (second.count(x) == 0) {
      ans = max(ans, x);
    }
  }
  cout << ans;
  return 0;
}

```