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.

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.

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

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： n个数字 求任意组合求和后为2的倍数的最大值

题解：n个数求和 sum 同时找到最小的为奇数的值（排序处理）sum为偶数 直接输出 若为奇数 减去最小奇数 输出

 #include<bits/stdc++.h>

#define LL __int64

using namespace std;

int n;

LL a[100005];

LL sum=0;

int main()

{

    scanf("%d",&n);

    sum=0;

    for(int i=0;i<n;i++)

       {

        scanf("%I64d",&a[i]);

        sum+=a[i];

       }

    sort(a,a+n);

    for(int j=0;j<n;j++)

    {

        if(sum%2==0)

            break;

        if(a[j]%2==1)

        {

            sum=sum-a[j];

            break;

        }

    }

    printf("%I64d\n",sum);

    return 0;

}