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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<vector<int>> Graph;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
#define pi 3.14159265359
#define inf 2147483647
#define INF 9223372036854775807
#define mod 1000000007
#define mod2 998244353

int main() {
	int N; cin >> N;
	vector<ll> A(N), B(N);
	for(int i = 0; i < N; i++) {
		cin >> A[i];
		B[i] = A[i];
	}
	for(int i = 0; i < N - 1; i++) {
		if(A[i] < 0) {
			if(i == N - 2 && A[N - 1] > 0 && -A[N - 2] < A[N - 1]) continue;
			A[i] *= -1; A[i + 1] *= -1;
		}
	}
	for(int i = N - 2; i >= 0; i--) {
		if(B[i + 1] < 0) {
			if(i == 0 && B[0] > 0 && B[0] > -B[1]) continue;
			B[i] *= -1; B[i + 1] *= -1;
		}
	}
	ll sum = 0, mus = 0;
	for(int i = 0; i < N; i++) sum += A[i];
	for(int i = 0; i < N; i++) mus >= B[i];
	ll ans = max(sum, mus);
	cout << ans << endl;
	return 0;
}