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: Accepted)

#include<bits/stdc++.h>
#define pb push_back
#define int long long
#define rint register int
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=1e5+10;
const int inf=0x3f3f3f3f;

int n,cnt,ans;
int arr[maxn];

signed main(){
	scanf("%lld",&n);
	for(rint i=1;i<=n;++i) scanf("%lld",&arr[i]);
	for(rint i=1;i<=n;++i) if(arr[i]<0) ++cnt;
	for(rint i=1;i<=n;++i) ans+=abs(arr[i]);
	if(cnt%2==1){
		int mn=inf;
		for(rint i=1;i<=n;++i) mn=min(mn,abs(arr[i]));
		ans-=2*(abs(mn));
	}
	printf("%lld\n",ans);
	return 0;
}