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<cstdio>
#include<cstring>
#include <cstdlib>  
#include <cmath>   
#include<cctype>
#include<string>
#include<set>
#include<iomanip>
#include <map>
#include<algorithm>
#include <functional>
#include<vector>
#include<climits>
#include<stack>
#include<queue>
#include <deque>
#include <climits>
#include <typeinfo>
#include <utility> 
#define all(x) (x).begin(),(x).end()
#define rep(i,m,n) for(int i = m;i < n;++i)
#define pb push_back
#define fore(i,a) for(auto &i:a)
#define rrep(i,m,n) for(int i = m;i >= n;--i)
#define INF INT_MAX/2
using namespace std;
using ll = long long;
using R = double;
using Data = pair<ll, vector<int>>;
const ll MOD = 1e9 + 7;
const ll inf = 1LL << 50;
struct edge { ll from; ll to; ll cost; };

ll dp[101010];

int main() {
	ll n;
	cin >> n;

	vector<ll>a(n);
	rep(i, 0, n)cin >> a[i];
	
	if (n % 3 == 0) {
		ll ans = 0;
		rep(i, 0, n) {
			ans += abs(a[i]);
		}
		cout << ans << endl;
	}
	else {
		ll ans = 0;
		ll mi = inf;
		rep(i,0,n){
			ans += abs(a[i]);
			mi = min(mi,abs(a[i]));
		}
		ans -= mi;
		cout << ans << endl;
	}


	return 0;
}