B. Odd sum

B. Odd sum

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given sequence a1, a2, ..., an of integer numbers of length n. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.

Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

You should write a program which finds sum of the best subsequence.

Input

The first line contains integer number n (1 ≤ n ≤ 105).

The second line contains n integer numbers a1, a2, ..., an ( - 104 ≤ ai ≤ 104). The sequence contains at least one subsequence with odd sum.

Output

Print sum of resulting subseqeuence.

Examples

input

4-2 2 -3 1

output

3

input

32 -5 -3

output

-1

Note

In the first example sum of the second and the fourth elements is 3.

数组排序一下，同时可以想到，该序列最大的子序列和一定是所有正数的和，要找最大子序列的奇数和，只需要判断该所有正数的和，如果为奇数，直接输出，否则找到绝对值最小的一个奇数，然后用和减去该数即可；

代码如下：

////  main.cpp//  B. Odd sum////  Created by 徐智豪 on 2017/4/15.//  Copyright © 2017年 徐智豪. All rights reserved.//#include <iostream>#include <algorithm>#include <cmath>#include <stdio.h>#include <queue>using namespace std;int main(int argc, const char * argv[]) {    priority_queue<int> jishu;    int n;    cin>>n;    int x;    long long maxsum=0;    for(int i=1;i<=n;i++)    {        cin>>x;        if(x>=0)        {            maxsum+=x;            if(x%2)                jishu.push(-x);        }       else       {           if(x%2)               jishu.push(x);       }    }    if(maxsum%2)        cout<<maxsum<<endl;    else cout<<maxsum+jishu.top()<<endl;        return 0;}