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.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class Solution {

  static final boolean DEBUG = false;

    BufferedReader in;
    StringTokenizer st;
    PrintWriter out;

    void solve() throws IOException {
    int n = ni();
    int fives = 0, zeroes = 0;
    for (int i = 0; i < n; ++i) {
      if (ni() == 5) {
        ++fives;
      } else {
        ++zeroes;
      }
    }
    int len = fives / 9;
    if (zeroes == 0) {
      out.println(-1);
      return ;
    }
    for (int i = 0; i < len * 9; ++i) {
      out.print(5);
    }
    if (len == 0) {
      zeroes = 1;
    }
    for (int i = 0; i < zeroes; ++i) {
      out.print(0);
    }
    }

  void dbg(Object ... objs) {
    if (!DEBUG) {
      return ;
    }
    for (Object o : objs) {
      String printLine;
      if (o.getClass().isArray()) {
        printLine = arrayToString(o);
      } else {
        printLine = o.toString();
      }
      System.err.print(printLine + " ");
    }
    System.err.println();
  }

  String arrayToString(Object o) {
    if (o instanceof long[]) 
      return Arrays.toString((long[]) o);
    if (o instanceof int[])
      return Arrays.toString((int[]) o);
    if (o instanceof short[])
      return Arrays.toString((short[]) o);
    if (o instanceof char[])
      return Arrays.toString((char[]) o);
    if (o instanceof byte[])
      return Arrays.toString((byte[]) o);
    if (o instanceof double[])
      return Arrays.toString((double[]) o);
    if (o instanceof float[])
      return Arrays.toString((float[]) o);
    if (o instanceof boolean[])
      return Arrays.toString((boolean[]) o);
    if (o instanceof Object[])
      return Arrays.deepToString((Object[]) o);
    throw new IllegalStateException();
  }
  

    public void run() throws IOException {
        Locale.setDefault(Locale.US);
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }

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

    int ni() throws IOException {
        return Integer.valueOf(ns());
    }

    long nl() throws IOException {
        return Long.valueOf(ns());
    }

    double nd() throws IOException {
        return Double.valueOf(ns());
    }

    public static void main(String[] args) throws IOException {
        new Solution().run();
    }

  class Timer {
    
    long time;
    
    void start() {
      time = System.currentTimeMillis();
    }
    long time() {
      return System.currentTimeMillis() - time;
    }
    void print() {
      print("Time spent = ");
    } 
    
    void print(String message) {
      dbg(message, time());
    }

  }
}
