MASUM - Maximum Sum of the Array

You are given an array of N integers. You tried to sum all the elements of the array, but you see the summation does not reach the maximum.

So you decide to divide the array into two arrays. You can choose any index ( 1 ≤ i ≤ N ) from array Ar and remove the value from the Ar array and add it to another array A. Also, you can choose any index ( 1 ≤ j ≤ N ) and remove the value from Ar and add it to another array B

Suppose an array ar = [ 1, 2, 3, 0, 5 ], so you can choose index 1, 3, 5 and remove it from array ar and add it to array A. So ar = [ ., 2, ., 0, . ] and A=[ 1, 3, 5 ]. You can choose index 2, 4 and remove it from array ar and add it to array B. So ar = [ ., ., ., ., . ] and B = [ 2, 0 ].

You can do any operation until array ar becomes empty.

Here is the main problem. You need to divide ar in such way that SUM(A) - SUM(B) gets maximized.
Here SUM(X) means summation of the array x. Example x = [ 1, 5, 8] so SUM(x) = 1 + 5 + 8 = 14.
Now Print the maximum SUM(A) - SUM(B) you can get.
Both A and B array needs to contain at least 1 element from array ar.

Input

  • First Line will contain N ( 2 ≤ N ≤ 106), the number of elements present in the array.
  • Second line will contain array ar of N elements. ( -109 ≤ ari ≤ 109 ).

Output

Print the maximum value you can get of SUM(A) - SUM(B) after doing those operations.

Example

Input 01:
5
1 2 3 4 5

Output 01:
13
Input 02:
6
-1 1 2 -2 3 -3

Output 02:
12

Added by:Efat Shikder
Date:2023-07-02
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All

© Spoj.com. All Rights Reserved. Spoj uses Sphere Engine™ © by Sphere Research Labs.