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: Accepted)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define Khela ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ll long long int
#define llu long long unsigned int
#define pf printf
#define sf scanf
#define f first
#define s second
#define pb push_back
#define mk make_pair
#define pii pair<int ,int>
 
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1,-1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
 
typedef tree<pair<int,int>,null_type,less<pair<int,int>>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
int Set(int N,int pos){return N=N | (1<<pos);}
int reset(int N,int pos){return N= N & ~(1<<pos);}
bool check(int N,int pos){return (bool)(N & (1<<pos));}
const int Max=1e9+100;
const ll Maxx=1e18+500;
const int sz=1e3+100;
 
int main()
{
    Khela
    ll a,b,c,i,j,k,q,p,x,y,ct,ct1,m,l,r,x1,y1,mn,h,sum1,in,z,mid,n,mx;
    char ch;
    double d;
    string str1,str2,str;
    bool bl,bl1;
    int t,cs=1;
    while(cin>>n)
    {
        ll ara[n+2];
        for(i=0;i<n;i++) cin>>ara[i];
        ll last=0;
        ara[n]=0ll;
        ll sum=0;
        ll mn=Maxx;
        ll ct=0;
        for(i=0;i<=n;i++){
            if(ara[i]>=0){
                if(ct>0){
                    if((ct+last)%2==0) last=0;
                    else last=1;
                }
                ct=0;
            }
            else{
                ++ct;
            }
            sum+=abs(ara[i]);
            if(i<n) mn=min(mn,abs(ara[i]));
        }
        if(last>0) sum-=(2*mn);
        cout<<sum<<endl;
    } 
 
 
 
 
 
}