【CodeForces】#621A - Wet Shark and Odd and Even（水）

题目链接：点击打开题目

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：让给出的数的和为偶数，并且使之最大。

思路：在输入阶段统计所有的和，并更新最小的奇数，如果最后和为偶数的话，直接输出；如果和为奇数，减去最小的奇数即可。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
	int n;
	__int64 sum = 0;
	int mi = 0x7fffffff;
	scanf ("%d",&n);
	while (n--)
	{
		int t;
		scanf ("%d",&t);
		sum += t;
		if (t & 1)
			mi = min (t , mi);
	}
	if (sum & 1)
		printf ("%I64d\n",sum - mi);
	else
		printf ("%I64d\n",sum);
	return 0;
}