#include <bits/stdc++.h>
using namespace std;
long long POW(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b & 1) return (a * POW(a, b - 1)) % 1000000007LL;
  long long t = POW(a, b / 2);
  return (t * t) % 1000000007LL;
}
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int dx[] = {0, -1, 1, 0}, dy[] = {1, 0, 0, -1};
int ddx[] = {0, 0, 1, 1, 1, -1, -1, -1}, ddy[] = {1, -1, 1, 0, -1, 1, 0, -1};
int main() {
  int ans = -1987654321;
  int n;
  scanf("%d", &n);
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) {
    int x;
    scanf("%d", &x);
    if (x < 0)
      ans = max(ans, x);
    else {
      int t = sqrt(x);
      if (t * t != x) ans = max(ans, x);
    }
  }
  printf("%d", ans);
}

### Prompt

Please formulate 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 POW(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b & 1) return (a * POW(a, b - 1)) % 1000000007LL;
  long long t = POW(a, b / 2);
  return (t * t) % 1000000007LL;
}
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int dx[] = {0, -1, 1, 0}, dy[] = {1, 0, 0, -1};
int ddx[] = {0, 0, 1, 1, 1, -1, -1, -1}, ddy[] = {1, -1, 1, 0, -1, 1, 0, -1};
int main() {
  int ans = -1987654321;
  int n;
  scanf("%d", &n);
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) {
    int x;
    scanf("%d", &x);
    if (x < 0)
      ans = max(ans, x);
    else {
      int t = sqrt(x);
      if (t * t != x) ans = max(ans, x);
    }
  }
  printf("%d", ans);
}

```