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>

using namespace std;

#define rep(i,n) for(int i=0 ; i<(n) ; i++)
#define repr(i,n) for(int i=(n) ; i>=0 ; i--)
#define FOR(i,m,n) for(int i=(m) ; i<(n) ; i++)
#define FORR(i,m,n) for(int i=(m) ; i>=(n) ; i--)

#define All(x) (x).begin(),(x).end()

typedef long long ll;
typedef unsigned long long ull;

const int INF = 1e9;

int main(){
   int n;cin >> n;
   vector<int> a(n);
   rep(i,n){
      cin >> a[i];
   }
   sort(All(a));
   int minus = 0;
   int minNum = 0;
   for(int i=0;i<n;i++){
      if(a[i]>=0){
         minNum = i;
         break;
      }
      minus++;
   }
   ll ans = 0;
   if(minus%2 == 0){
      rep(i,n){
         ans+= abs(a[i]);
      }
      cout << ans << endl;
   }else{
      for(int i=0;i<n;i++){
         ans+= abs(a[i]);
      }
      cout << ans - min(abs(a[minNum]),abs(a[minNum-1]))*2 << endl;
   }
   return 0;
}