CodeForces352A Jeff and Digits

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, ..., a_n (a_i = 0 or a_i = 5). Number a_i 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.

Example

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.

能被90整除也就是说可以被10和9整除。

对于前者也就是说末尾一定是0。

对于后者就是5的个数一定是9的倍数。

嗯于是代码写矬辣QAQ

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <cmath>
#include <cctype>
#include <string>
#include <cfloat>
#include <stack>
#include <cassert>

using namespace std;

int num0, num5, n;

int main() {
    scanf("%d", &n);
    for(int i = 1, x ; i <= n ; i ++) {
        scanf("%d", &x);
        if(x == 0) {
            num0 ++;
        } else {
            num5 ++;
        }
    }
    if(num0 == 0) puts("-1");
    else {
        if(num5 < 9) {
            puts("0");
        } else {
            for(int i = num5 ; i >= 0 ; i --){
                if(i % 9 == 0) {
                    for(int j = 1 ; j <= i ; j ++) {
                        putchar('5');
                    }
                    break;
                }
            }
            for(int i = 1 ; i <= num0 ; i ++) {
                putchar('0');
            }
        }
    }
}