【CodeForces 797B】Odd sum（模拟）



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, ..., a_n 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, ..., a_n ( - 10⁴ ≤ a_i ≤ 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.

题目大意：找一个子序列，使其和为奇数并且和最大

思路：虽然是子序列，但与顺序无关，只有奇数+偶数才能等于奇数，题目保证有解，则至少有1个奇数，将所有正偶数都加起来，奇数递减排序，第一个奇数一定要加上，现在sum是奇数，之后只能加偶数，两个奇数的和是偶数，依次加上相邻两个和为正的奇数

#include <bits/stdc++.h>#define manx 100005using namespace std;bool cmp(int a,int b){    return a > b;}int main(){    int a[manx],l,n,x;    while(~scanf("%d",&n)){        memset(a,0,sizeof(a));        l=0;        int sum=0;        for (int i=0; i<n; i++){            scanf("%d",&x);            if (x%2) a[l++]=x; //odd            else if (x > 0) sum+=x;        }        sort(a,a+l,cmp);        sum+=a[0];        for (int i=1; i<l-1; i+=2){            if (a[i] + a[i+1] >= 0) sum+=a[i]+a[i+1];        }        printf("%d\n",sum);    }    return 0;}