Codeforces Round #341 (Div. 2)A

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.

Sample test(s)

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： 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
",sum);
    return 0;
}