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.
from sys  import stdin,stdout
 
import bisect

import math

from collections import Counter
 
def st():
    return list(stdin.readline().strip())
 
def inp():
    return int(stdin.readline())
 
def li():
    return list(map(int,stdin.readline().split()))
 
def mp():
    return map(int,stdin.readline().split())
 
def pr(n):
    stdout.write(str(n)+"\n")
 
def soe(limit):
    l=[1]*(limit+1)
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0
 
    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime
 
def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)
 
def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
            n=n-1
        else:
            a=(a*a)
            n=n>>1
    return r
    
def solve():
    n=inp()
    l=li()
    x=[]
    for i in l:
        if i>0:
            if i**0.5 != math.floor(i**0.5):
                x.append(i)
        elif i<0:
            x.append(i)
    pr(max(x))
            
            
        
            

for _ in range(1):
    solve()
