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>
using namespace std;
#define inf 1000000000
#define INF 1000000000000000
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define PLL pair<ll,ll>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
#define IN(a,n) rep(i,n){ cin>>a[i]; }
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define PI 3.14159265
#define F first
#define S second
#define PB push_back
#define EB emplace_back
void init(){
  cin.tie(0);
  ios::sync_with_stdio(false);
}


int n;
ll a[110000];


int main(){
  cin>>n;
  int r=0,res=0;
  rep(i,n){
    cin>>a[i];
    if(a[i]<0) res++;
    if(a[i]==0) r++;
  }

  if(res%2==0||(res+r)%2==0){
    ll ans=0;
    rep(i,n){
      ans+=abs(a[i]);
    }
    cout<<ans<<endl;
  } else{
    int s=0;

    rep(i,n){
      if(abs(a[i])<abs(a[s]))
        s=i;
    }

    ll ans=-abs(a[s]);
   
    rep(i,n){
      if(i==s) continue;
      ans+=abs(a[i]);
    }

    cout<<ans<<endl;
  }

}