Wet Shark and Odd and Even



 Wet Shark and Odd and Even

Crawling in process...Crawling failedTime Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatusPractice 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 then 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

31 2 3

Output

6

Input

5999999999 999999999 999999999 999999999 999999999

Output

3999999996

Hint

In the first sample, we can simply take all three integers for a total sum of6.

In the second sample Wet Shark should take any four out of five integers999 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;}