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.
import java.util.*;
public class PerfectSquares {

	private static final double Ep = Math.pow(10, -9);
	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		ArrayList<Integer> a = new ArrayList<Integer>();
		for(int i = 0; i < n; i++) {
			a.add(sc.nextInt());
		}
		a.sort(Collections.reverseOrder());
		for(int i = 0; i < n; i++) {
			int key=a.get(i);
//			System.out.println(Math.sqrt(16));
//			System.out.println(Math.sqrt(64));
//			System.out.println(Math.sqrt(56));
//			System.out.println(isSquare(key));
			if(!isSquare(key)) {
				System.out.println(key);
				break;
			}
		}
		sc.close();
		
	}

	private static boolean isSquare(int key) {
		
		for(int i = 0; i < (int)Math.sqrt(key)+1; i++) {
//			System.out.println(i*i-Ep);
//			System.out.println( i*i+Ep);
//			System.out.println(key);
			if(key == i*i) {
				return true;
			}
		}
		return false;
	}
	
}
