Jeff's got n cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got?

Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.

Input

The first line contains integer n (1 ≤ n ≤ 103). The next line contains n integers a1, a2, ..., an (ai = 0 or ai = 5). Number ai represents the digit that is written on the i-th card.

Output

In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.

Examples

Input

4
5 0 5 0


Output

0


Input

11
5 5 5 5 5 5 5 5 0 5 5


Output

5555555550

Note

In the first test you can make only one number that is a multiple of 90 — 0.

In the second test you can make number 5555555550, it is a multiple of 90.
//package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class D {

    public static void main(String[] args) throws IOException
    {
	Scanner bf = new Scanner(System.in);
	PrintWriter out = new PrintWriter(System.out);
	int n = bf.nextInt();
	int zero = 0, five = 0;
	for (int i = 0; i < n; i++)
	{
	    int num = bf.nextInt();
	    if(num == 0) zero++;
	    if(num == 5) five++;
	}
	if(zero == 0)
	    out.println(-1);
	else{
	    StringBuilder sb = new StringBuilder();
	    int size = five/9*9;
	    for (int i = 0; i < size; i++)
	    {
		sb.append("5");
	    }
	    if(size > 0){
		for (int i = 0; i < zero-1; i++)
		{
		    sb.append("0");
		}
	    }
	    sb.append("0");
	    out.println(sb);
	}

	out.flush();
	out.close();
    }

    static class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s)
	{
	    br = new BufferedReader(new InputStreamReader(s));
	}

	public String next() throws IOException
	{
	    while (st == null || !st.hasMoreTokens())
		st = new StringTokenizer(br.readLine());
	    return st.nextToken();
	}

	public int nextInt() throws IOException
	{
	    return Integer.parseInt(next());
	}

	public long nextLong() throws IOException
	{
	    return Long.parseLong(next());
	}

	public String nextLine() throws IOException
	{
	    return br.readLine();
	}

	public double nextDouble() throws IOException
	{
	    String x = next();
	    StringBuilder sb = new StringBuilder("0");
	    double res = 0, f = 1;
	    boolean dec = false, neg = false;
	    int start = 0;
	    if (x.charAt(0) == '-')
	    {
		neg = true;
		start++;
	    }
	    for (int i = start; i < x.length(); i++)
		if (x.charAt(i) == '.')
		{
		    res = Long.parseLong(sb.toString());
		    sb = new StringBuilder("0");
		    dec = true;
		} else
		{
		    sb.append(x.charAt(i));
		    if (dec)
			f *= 10;
		}
	    res += Long.parseLong(sb.toString()) / f;
	    return res * (neg ? -1 : 1);
	}

	public boolean ready() throws IOException
	{
	    return br.ready();
	}
    }

}
