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 <cstring>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
#include <bitset>
#include <cstdlib>
#include <cmath>
#define INF 100000000000
#define MOD 1000000007
#define ll long long int
using namespace std;



int main(){
  ll N; cin>>N;
  vector <ll> A; A.resize(N+1,0);
  vector <ll> dp; dp.resize(N+1,0); dp[0] = 0;
  vector <ll> rp; rp.resize(N+1,0); rp[0] = 0;
  for(int i=0;i<N;i++) cin>>A[i+1];
  dp[1] = (A[1]);
  rp[1] = -(A[1]);

  //cout<<1<<endl;
  //cout<<dp[1]<<"   "<<rp[1]<<endl;

  for(int i=2;i<=N-1;i++){ // 漸化式
//  ll a = A[i-1];
 // ll b = A[i];

  ll dp1,dp2,dp3,dp4;
  ll rp1,rp2,rp3,rp4;

//cout<<i<<endl;
  

  dp1 = max(dp[i-2] + A[i-1] + A[i], dp[i-2] - A[i-1] - A[i]); 
  //cout<<dp1<<endl;
  dp2 = dp[i-1] + A[i];
  dp3 = max(rp[i-2] - A[i-1] + A[i] , rp[i-2] + A[i-1] - A[i]);
  if(i==2) dp3 = 0;
  //cout<<dp2<<endl;
  //cout<<dp3<<endl;
  dp4 = rp[i-1] - A[i];
  //cout<<dp4<<endl;

  rp1 = max(rp[i-2] - A[i-1] - A[i] , rp[i-2] + A[i-1] + A[i]);
  rp2 = rp[i-1] + A[i];
  rp3 = max(dp[i-2] + A[i-1] - A[i] , dp[i-2] - A[i-1] + A[i]);
  if(i==2) rp3 = 0;
  rp4 = dp[i-1] - A[i];
  
ll res_dp = max(dp1,dp2);
res_dp = max(res_dp,dp3);
dp[i] = max(res_dp , dp4);

ll res_rp = max(rp1,rp2);
res_rp = max(res_rp,rp3);
rp[i] = max(res_rp , rp4);


//cout<<dp[i]<<"   "<<rp[i]<<endl;


  }


cout<<max(dp[N-1] + A[N] , rp[N-1] - A[N]) <<endl;

  return 0;
}