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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

function Main(input){
 input = input.split("\n");
 inputStr = input[0].split(" ");
 var N = Number.parseInt(inputStr[0]);
 var A = input[1].split(" ");
 var i = 0;
 var out = 0;
 var sum = 0;
 var abssum = 0;
 var mainus_cnt = 0;
  
 //console.log(N);
 //console.log(A);  
  
 //負の数が幾つあるか数える
 for ( i = 0 ; i < N; i++) {
   if ( Number.parseInt(A[i]) < 0 ) {
     mainus_cnt += 1;
     if ( i > 0 && Number.parseInt(A[i-1]) === 0 ) {
       mainus_cnt = 0;
     }
   }
   if ( Number.parseInt(A[i]) === 0 ) {
     mainus_cnt = 0;
   }
 }
 //console.log(mainus_cnt);
  
 //文字列Aの最後と、そのひとつ前のどちらが絶対値として大きいか判定
 if ( ( mainus_cnt % 2 ) !== 0 ) {
   if ( Math.abs(Number.parseInt(A[N-1])) >= Math.abs(Number.parseInt(A[N-2]))) {
     out = N-2;
   } else {
     out = N-1;
   }
 }
 //console.log(out);
  
 //forで回しながら絶対値を足しこむ。
 for ( i=0 ; i<N ; i++) {
   //console.log(i);
   if ( out === 0 ) {
     abssum = abssum + Math.abs(Number.parseInt(A[i]));
   } else {
     if ( i === out) {
       abssum = abssum - Math.abs(Number.parseInt(A[i]));
     } else {
       abssum = abssum + Math.abs(Number.parseInt(A[i]));
     }
   }
   //sum = sum + Math.abs(Number.parseInt(A[i]));
 }
 console.log(abssum);
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));