Codeforces Round #341 (Div. 2)A Wet Shark and Odd and Even

思路：水题，直接全部加起来，并且维护一个最小的奇数，因为如果最后结果是奇数的话那么减去那个最小的奇数就会变成最大的偶数啦

#include<bits\stdc++.h>
using namespace std;
#define LL long long
int main()
{
    int n;
	cin >> n;
	LL ans = 0;
	LL mins = 1e9;
	while(n--)
	{
		LL temp;
		scanf("%lld",&temp);
		ans+=temp;
		if(temp&1)
			mins = min(temp,mins);
	}
	if(ans&1)
		ans-=mins;
	cout << ans << endl;
}

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.