CF 131 div2 B

B. Hometask

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?

You are given a set of digits, your task is to find the maximum integer that you can make from these digits. The made number must be divisible by 2, 3, 5 without a residue. It is permitted to use not all digits from the set, it is forbidden to use leading zeroes.

Each digit is allowed to occur in the number the same number of times it occurs in the set.

Input

A single line contains a single integer n (1 ≤ n ≤ 100000) — the number of digits in the set. The second line contains n digits, the digits are separated by a single space.

Output

On a single line print the answer to the problem. If such number does not exist, then you should print -1.

Sample test(s)

input

1
0

output

0

input

11
3 4 5 4 5 3 5 3 4 4 0

output

5554443330

input

8
3 2 5 1 5 2 2 3

output

-1

Note

In the first sample there is only one number you can make — 0. In the second sample the sought number is 5554443330. In the third sample it is impossible to make the required number.

这题看似简单，非常容易漏掉情况没有考虑。如果余1，可以删一个余数1的，或者两个余数2的！如果余2，可以删一个余数2的，或两个余数1的。最后注意！如果只有0出现的话，最后只输出一个0！所以要特判。

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<string>
#include<map>
using namespace std;
typedef long long LL;
const int maxn = 100 + 5;
const int INF = 1000000000;

int num[10];

int main(){
    int n;
    while(cin >> n){
        memset(num,0,sizeof(num));
        int total = 0;
        for(int i = 0;i < n;i++){
            int tem;
            cin >> tem;
            num[tem]++;
            total += tem;
        }
        total = total % 3;
        int tag = 1;
        if(total == 1){
            tag = 0;
            for(int i = 0;i < 10;i++){
                if(i%3 == 1 && num[i] > 0){
                    tag = 1;
                    num[i]--;
                    break;
                }
            }
            int cnt = 2;
            if(tag == 0){
                for(int i = 0;i < 10;i++){
                    while(i%3 == 2 && num[i] > 0 && cnt > 0){
                        num[i]--;
                        cnt--;
                    }
                }
                if(cnt == 0) tag = 1;
            }
        }
        else if(total == 2){
            int cnt = 2;
            tag = 0;
            for(int i = 0;i < 10;i++){
                if(i%3 == 2 && num[i] > 0){
                    num[i]--;
                    tag = 1;
                    break;
                }
            }
            if(tag == 0){
                for(int i = 0;i < 10;i++){
                    while(i%3 == 1 && num[i] > 0 && cnt > 0){
                        num[i]--;
                        cnt--;
                    }
                }
                if(cnt == 0) tag = 1;
            }
        }
        if(tag == 0 || num[0] == 0){
                cout << -1 << endl;
                continue;
        }
        else{
            int flag = 0;
            for(int i = 9;i >= 0;i--){
                if(i == 0 && flag == 0){
                    cout << 0 << endl;
                    break;
                }
                while(num[i] > 0){
                    if(i != 0) flag = 1;
                    num[i]--;
                    cout << i;
                }
            }
            cout << endl;
        }
    }
    return 0;
}