Recently, Bob has been thinking about a math problem.
There are N Digits, each digit is between 0 and 9. You need to use this N Digits to constitute an Integer.
This Integer needs to satisfy the following conditions:

• 1. must be an odd Integer.
• 2. there is no leading zero.
• 3. find the biggest one which is satisfied 1, 2.

Example:
There are three Digits: 0, 1, 3. It can constitute six number of Integers. Only "301", "103" is legal, while "130", "310", "013", "031" is illegal. The biggest one of odd Integer is "301".

There are multiple test cases. Please process till EOF.
Each case starts with a line containing an integer N ( 1 <= N <= 100 ).
The second line contains N Digits which indicate the digit
a1,a2,a3,⋯,an.(0≤ai≤9).

The output of each test case of a line. If you can constitute an Integer which is satisfied above conditions, please output the biggest one. Otherwise, output "-1" instead.

3
0 1 3
3
5 4 2
3
2 4 6

301
425
-1
/*
HDU 5055 数学题 
主要注意几点：
1、n=1的情况， 
2、没有奇数 -1 
3、n>1时，只有一个奇数，其他都为0  -1 
*/
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define N 101
int a[N];
int main()
{
	int n,i,oddNum,num,min;
	while(scanf("%d",&n)!=EOF)
	{
		oddNum=0;//奇数个数 
		num=0;//非0的个数 
		
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			if(a[i]&1)
				oddNum++;
			if(a[i]!=0)
				num++;
		}
		if(n==1)//n==1的判断 
		{
			if(a[0]&1)
				printf("%d\n",a[0]);
			else
				printf("-1\n");
			continue;
		}	
				
		if(oddNum<1||(oddNum==1&&num==1))
		{
			printf("-1\n");
			continue;
		}
		sort(a,a+n);//从小到大 
		for(i=0;i<n;i++)
			if(a[i]&1)//寻找最小的一个奇数放在最后 
			{
				min=i;
				break;
			} 
		for(i=n-1;i>=0;i--)
		{
			if(i!=min)
				printf("%d",a[i]);
		}
		printf("%d\n",a[min]);
	}
	return 0;
}