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<bits/stdc++.h>
#define INF 1e9
#define llINF 1e18
#define MOD 1000000007
#define pb push_back
#define mp make_pair 
#define F first
#define S second
#define ll long long
#define ull unsigned long long
#define vi vector<ll>
#define vvi vector<vi>
#define DBG_N(hoge) cerr<<" "<<(hoge)<<endl;
#define DBG cerr<<"!"<<endl;
#define BITLE(n) (1LL<<((ll)n))
#define BITCNT(n) (__builtin_popcountll(n))
#define SUBS(s,f,t) ((s).substr((f)-1,(t)-(f)+1))
#define ALL(a) (a).begin(),(a).end()
using namespace std;
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll n;cin>>n;
  vi A(n);
  bool fl=false;
  for(auto &a:A){
    cin>>a;
    if(a==0)fl=true;
  }
  ll ans=0;
  if(fl){
    for(int i=0;i<n-1;i++){
      ans+=abs(A[i]);
    }
  }else{
    ll minus=0;
    bool flfl=false;
    for(int i=0;i<n;i++){
      if(A[i]<0)minus++;

    }
    if(minus%2)flfl=true;
    if(flfl){
      for(int i=0;i<n;i++)A[i]=abs(A[i]);
      sort(ALL(A));
      for(int i=1;i<n;i++)
	ans+=A[i];
      ans-=A[0];
    }else{
      for(int i=0;i<n;i++)
	ans+=A[i];
    }
  }
  cout<<ans<<endl;
  return 0;
}