Wet Shark and Odd and Even


[align=center] Wet Shark and Odd and Even[/align]
Crawling in process...Crawling failedTime
Limit:2000MS    Memory Limit:262144KB    
64bit IO Format:%I64d & %I64u

SubmitStatus

Practice
CodeForces 621A

Description

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 by2) 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 integer0.

Input

The first line of the input contains one integer,
n (1 ≤ n ≤ 100 000). The next line containsn space separated integers given to Wet Shark. Each of these integers is in range from1
to 109, inclusive.

Output

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

Sample Input

Input

3
1 2 3

Output

6

Input

5
999999999 999999999 999999999 999999999 999999999

Output

3999999996

Hint

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.

#include<stdio.h>
#include<algorithm>
using namespace std;
long long a[100005];
int main()
{
int n;
scanf("%d",&n);
long long sum=0;
for(int i=0;i<n;i++)
{
scanf("%lld",&a[i]);
sum+=a[i];
}
sort(a,a+n);
for(int i=0;i<n;i++)
{
if(sum%2==0)
break;
else
{
if(a[i]%2!=0)
{
sum-=a[i];
}
}
}
printf("%lld\n",sum);
return 0;
}