Problem:
Score : 400 points

Problem Statement

There are N integers, A_1, A_2, ..., A_N, arranged in a row in this order.

You can perform the following operation on this integer sequence any number of times:

Operation: Choose an integer i satisfying 1 \leq i \leq N-1. Multiply both A_i and A_{i+1} by -1.

Let B_1, B_2, ..., B_N be the integer sequence after your operations.

Find the maximum possible value of B_1 + B_2 + ... + B_N.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

-10^9 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible value of B_1 + B_2 + ... + B_N.

Sample Input 1

3
-10 5 -4

Sample Output 1

19

If we perform the operation as follows:

Choose 1 as i, which changes the sequence to 10, -5, -4.

Choose 2 as i, which changes the sequence to 10, 5, 4.

we have B_1 = 10, B_2 = 5, B_3 = 4. The sum here, B_1 + B_2 + B_3 = 10 + 5 + 4 = 19, is the maximum possible result.

Sample Input 2

5
10 -4 -8 -11 3

Sample Output 2

30

Sample Input 3

11
-1000000000 1000000000 -1000000000 1000000000 -1000000000 0 1000000000 -1000000000 1000000000 -1000000000 1000000000

Sample Output 3

10000000000

The output may not fit into a 32-bit integer type.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <cstdint>
#include <sstream>
#include <map>
#include<vector>

using namespace std;
typedef long long ll;

ll ans,mx,sum,mn=1e18,cnt,flag;

ll b[100000],idx;
int main(){
	ll n;
	cin>>n;
	ll a[n];
	for(ll i=0; i<n; i++){
		cin>>a[i];
		if(a[i]<=0) b[idx++]=i;
	}
	for(ll i=0; i<n; i++){
		if(a[i]<0){
			bool ok=false;
			for(ll j=0; j<idx; j++){
				if(b[j]>i){
					a[i]=abs(a[i]);
					a[b[j]]=abs(a[b[j]]);
					ok=true;
					break;
				}
			}
			if(ok==false){
				for(ll j=i; j<n; j++){
					if(abs(a[j])<mn){
						flag=j;
						mn=abs(a[j]);
					}
				}
				a[i]=abs(a[i]);
				a[flag] *=-1;
			}
		}
	}
	for(ll i=0; i<n; i++){
		sum+=a[i];
	}
	cout<<sum;
}