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 5.4.1) (judge verdict: Wrong Answer)

// AtCoder ABC125: D - Flipping Signs
// 2019.9.9 bal4u

#include <stdio.h>

typedef long long ll;

#if 1
int getchar_unlocked(void);
#define gc() getchar_unlocked()
#else
#define gc() getchar()
#endif

int in() {  // 整数の入力
	int n = 0, c = gc();
	if (c == '-') {	c = gc();
		do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
		return -n;
	}
	do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
	return n;
}

#define ABS(a) ((a)>=0?(a):-(a))
#define MAX(a,b) ((a)>=(b)?(a):(b))
#define MIN(a,b) ((a)<=(b)?(a):(b))

int N;
int A[100005];

ll calc()
{
	int i, minus, a, b; ll ans;
	
	minus = 0, ans = 0;
	for (i = 1; i <= N-2; i++) {
		if (A[i] == 0) minus = 0;
		else {
			if (minus) {
				if (A[i] < 0) minus = 0, ans -= A[i];
				else ans += A[i];
			} else {
				if (A[i] < 0) minus = 1, ans -= A[i];
				else ans += A[i];
			}
		}
	}
	a = A[N-1], b = A[N];
	if (minus) a = -a;
	if (a < 0 && b < 0 || a >= 0 && b >= 0) ans += ABS(a) + ABS(b);
	else {
		int aa, bb;
		aa = MAX(ABS(a), ABS(b)), bb = MIN(ABS(a), ABS(b));
		ans += aa - bb;
	}
	return ans;
}

int main()
{
	int i, j, t; ll a, ans;
	
	N = in(); for (i = 1; i <= N; i++) A[i] = in();
	ans = calc();
	
	for (i = 1, j = N; i < j; i++, j--) t = A[i], A[i] = A[j], A[j] = t;
	a = calc();
	
	if (a > ans) ans = a;
	printf("%lld\n", ans);
	return 0;
}