codeforce 621A（水题）

A. Wet Shark and Odd and Even

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.

Note, that if Wet Shark uses no integers from the n integers, the sum is an even integer 0.

Input

The first line of the input contains one integer, n (1 ≤ n ≤ 100 000). The next line contains n space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.

Output

Print the maximum possible even sum that can be obtained if we use some of the given integers.

Examples

input

3
1 2 3

output

6

input

5
999999999 999999999 999999999 999999999 999999999

output

3999999996

Note

In the first sample, we can simply take all three integers for a total sum of 6.

In the second sample Wet Shark should take any four out of five integers 999 999 999.

题意：求一组数的最大数，且这个最大数可以整除2 （这一组数全为正整数）

题解：求出所有数的和   和最小的奇数如果和为偶数直接输出，和为奇数减去最小的奇数后输出

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 100100
#define mod 10007
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
LL s[MAX];
int main()
{
    LL i,n,m,sum;
	while(scanf("%lld",&n)!=EOF)
	{
		sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%lld",&s[i]);
			sum+=s[i];
		}
		sort(s,s+n);
		if(sum&1)
		{
			for(i=0;i<n;i++)
			{
				if(s[i]&1)
				{
					sum=sum-s[i];
					break;
				}
			}
		}
		printf("%lld
",sum); 
	} 
	return 0;
}