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.
#include <bits/stdc++.h>
using namespace std;
long long int nCr1[1001][1001];
long long int nCr(long long int n, long long int r) {
  if (nCr1[n][r]) return nCr1[n][r];
  if (r == 1) return nCr1[n][r] = n;
  if (n == r) return nCr1[n][r] = 1;
  if (r == 0) return nCr1[n][r] = 1;
  return nCr1[n][r] = (nCr(n - 1, r) + nCr(n - 1, r - 1)) % 1000000007;
}
bool prime(long long int n) {
  long long int i;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, i, k;
  cin >> n;
  long long int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (i = n - 1; i >= 0; i--) {
    k = sqrt(a[i]);
    if (k * k != a[i]) {
      cout << a[i];
      break;
    }
  }
}
