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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3

N = int(input().split()[0])
a_list = list(map(int, input().split()))

odd_n = len([a for a in a_list if a < 0])

# 負の数の個数が偶数の場合は、全部正にできる
if odd_n % 2 == 0:
    s = 0
    for a in a_list:
        s += abs(a)
else:
    # 負の数の個数が奇数の場合は、0があったら全部正にできる
    # 0がない場合は、どれか1個が負のままで残るので、絶対値が一番小さいものを残す
    s = 0
    if 0 in a_list:
        for a in a_list:
            s += abs(a)
    else:
        min_abs = float('inf')
        for a in a_list:
            s += abs(a)
            min_abs = min(min_abs, abs(a))
        s -= min_abs * 2

ans = s
print(ans)