#include <bits/stdc++.h>
using namespace std;
long long gcd(long long n1, long long n2) {
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % 1000000007;
    base = (base * base) % 1000000007;
    exponent /= 2;
  }
  return ans;
}
bool isPerfectSquare(long long n) {
  long long squareRootN = (long long)round((sqrt(n)));
  if (squareRootN * squareRootN == n) {
    return true;
  } else {
    return false;
  }
}
int main() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (int i = n - 1; i >= 0; i--) {
    if (isPerfectSquare(a[i]) == false) {
      cout << a[i] << '\n';
      break;
    }
  }
}

### Prompt

Your challenge is to write a cpp solution 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;
long long gcd(long long n1, long long n2) {
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % 1000000007;
    base = (base * base) % 1000000007;
    exponent /= 2;
  }
  return ans;
}
bool isPerfectSquare(long long n) {
  long long squareRootN = (long long)round((sqrt(n)));
  if (squareRootN * squareRootN == n) {
    return true;
  } else {
    return false;
  }
}
int main() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (int i = n - 1; i >= 0; i--) {
    if (isPerfectSquare(a[i]) == false) {
      cout << a[i] << '\n';
      break;
    }
  }
}

```