#include <bits/stdc++.h>
using namespace std;
vector<int> a;
int n;
void inp() {
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  return;
}
bool check(int x) {
  for (int i = 0; i < 1001; i++)
    if (i * i == x) return false;
  return true;
}
void solve() {
  int ans = -int(1e7);
  for (int i = 0; i < n; i++)
    if (check(a[i])) ans = max(ans, a[i]);
  cout << ans << '\n';
  return;
}
int main() {
  clock_t beg = clock();
  {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
  };
  inp();
  solve();
  clock_t end = clock();
  fprintf(stderr, "%.3f sec\n", (double)(end - beg) / CLOCKS_PER_SEC);
  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;
vector<int> a;
int n;
void inp() {
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  return;
}
bool check(int x) {
  for (int i = 0; i < 1001; i++)
    if (i * i == x) return false;
  return true;
}
void solve() {
  int ans = -int(1e7);
  for (int i = 0; i < n; i++)
    if (check(a[i])) ans = max(ans, a[i]);
  cout << ans << '\n';
  return;
}
int main() {
  clock_t beg = clock();
  {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
  };
  inp();
  solve();
  clock_t end = clock();
  fprintf(stderr, "%.3f sec\n", (double)(end - beg) / CLOCKS_PER_SEC);
  return 0;
}

```