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: Time Limit Exceeded)

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define repi(i, a, b) for(int i=(a); i>(b); i--)
#define db(x) (cerr << #x << ": " << (x) << '\n')
#define sync ios_base::sync_with_stdio(false), cin.tie(NULL)
#define cps CLOCKS_PER_SEC
#define tests(t) int t; cin >> t; while(t--)
#define iceil(n, x) (((n) + (x) - 1) / (x))
#define ll long long
#define gcd __gcd
#define eb emplace_back
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define sz size()
#define all(v) (v).begin(), (v).end()
#define uni(v) sort(all(v)), (v).erase(unique(all(v)), (v).end())
#define pii pair<int, int>
#define vi vector<int>
#define vpii vector<pii>
#define vvi vector<vi>
#define fi first
#define se second
#define mt make_tuple
#define pqueue priority_queue
#define bitcount(x) __builtin_popcount(x)
#define PI acos(-1.0)
#define EPS 1e-9
#define mod 1000000007
using namespace std;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ','); cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}

template<typename T1, typename T2>
ostream& operator << (ostream& os, const pair<T1, T2>& p) { return os << '(' << p.fi << ", " << p.se << ')'; }

template<typename T>
void printv(const T& v) { for(auto i : v) cerr << i << ' '; cerr << '\n'; }

template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;

template<typename T>
using maxpq = priority_queue<T>;

//All indexing is 0-based
using namespace __gnu_pbds;
template<class key, class cmp = std::less<key>>
using ordered_set = tree<key, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
//methods: find_by_order(k); & order_of_key(k);
//To make it an ordered_multiset, use pairs of (value, time_of_insertion)
//to distinguish values which are similar

template<class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;

//Returns no. of values x for which ceil(n / x) == y (y must be > 1).
inline ll CC(ll n, ll y) { return iceil(n, y-1) - iceil(n, y); }

//Returns no. of values x for which floor(n / x) == y
inline ll FF(ll n, ll y) { return n / y - n / (y+1); }

//a and b are assumed to be taken modulo p
inline int add(int a, int b, int p = mod){ int c = a + b; if(c >= p) c -= p; return c; }
inline int sub(int a, int b, int p = mod){ int c = a - b; if(c < 0) c += p; return c; }
inline int mul(int a, int b, int p = mod){ return (a * 1ll * b) % p; }

#define N 100005
#define int ll
// #define trace(...) 42

const int inf = 1e18;

int n, a[N];
int dp[N][2];
bool vis[N][2];

int rec(int i, int j) {
    if(i == n-1) {
        return j ? -a[i] : a[i];
    }

    int& ans = dp[i][j];

    if(vis[i][j]) return ans;

    vis[i][j] = 0;

    ans = -inf;
    int k = j ? -a[i] : a[i];

    //Flip i and (i+1)
    ans = max(ans, -k + rec(i+1, 1));

    //Don't flip
    ans = max(ans, k + rec(i+1, 0));

    return ans;
}

main()
{   
    #ifdef CP
        freopen("/home/tarun/Desktop/input.txt", "r", stdin);
     // freopen("/home/tarun/Desktop/output.txt", "w", stdout);
    #endif
    sync;
    clock_t clk = clock();
    cerr << "I will return...\n";

    cin >> n;
    rep(i, 0, n) cin >> a[i];
    memset(vis, 0, sizeof vis);

    cout << rec(0, 0) << '\n';

    cerr << "...don't you ever hang your head.\n";
    cerr << "Time (in ms): " << (double)(clock() - clk) * 1000.0 / cps << '\n';
}

//Compile using:
//g++ -o filename.exe --std=c++11 filename.cpp
//Use -D CP for defining a macro CP in the local environment