Dessert Wizard Problem

Dessert Wizard Problem code: DELISH

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It's finally summer in Chefland! So our chef is looking forward to prepare some of the best "beat-the-heat" dishes to attract more customers. He summons the Wizard of Dessert to help him with one such dish.

The wizard provides the chef with a sequence of N ingredients where the i^th ingredient has a delish value of D[i]. The preparation of the dish takes place in two phases.

Phase 1 : The chef chooses two indices i and j and adds the ingredients i, i+1, ..., j to his dish. He also finds the sum of the delish value in this range i.e D[i] + D[i+1] + ... + D[j].

Phase 2 : The chef chooses two more indices k and l and adds the ingredients k, k+1, ..., l to his dish. He also finds the sum of the delish value in this range i.e D[k] + D[k+1] + ... + D[l].


Note that 1 ≤ i ≤ j < k ≤ l ≤ N.

The total delish value of the dish is determined by the absolute difference between the values obtained in the two phases. Obviously, the chef wants to maximize the total delish value for his dish. So, he hires you to help him.

Input

First line of input contains an integer T denoting the number of test cases. For each test case, the first line contains an integer N denoting the number of ingredients. The next line contains N space separated integers where the i^th integer represents the delish value D[i] of the i^th ingredient.

Output

Print the maximum delish value of the dish that the chef can get.

Constraints

• 1 ≤ T ≤ 50
• 2 ≤ N ≤ 10000
• -1000000000 (−10⁹) ≤ D[i] ≤ 1000000000 (10⁹)

Example

Input:
2
5
1 2 3 4 5
4
1 1 -1 -1

Output:
13
4

Explanation

Example case 1.

Chef can choose i = j = 1, k = 2, l = 5.

The delish value hence obtained is | (2+3+4+5) − (1) | = 13 .

Example case 2.

Chef can choose i = 1, j = 2, k = 3, l = 4.

The delish value hence obtained is | ( ( −1 ) + ( −1 ) ) − ( 1 + 1 ) | = 4 .

/*思路：分别求得从0-n序列的最大值和最小值，保存在max1[] , min2[]内，
    再求得n-0序列的最大值和最小值，保存在max2[] , min1[]内
最后只要找到max{abs(max1[] - min1[] ) , (max2[] - min2[] )}即可;

*/
#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std ;
#define MAX 10004
#define inf 1000000000
long long int min1[MAX] , min2[MAX] , max1[MAX] , max2[MAX] , a[MAX];

/*求0-n之间的最大值*/
void find_Max(int n )
{
    long long int i , sum = 0 , m = -inf , sum1 = 0 , m1 = inf ;
    for( i = 0 ; i < n ; i ++ )
    {

        sum1 += a[i] ;
        if(sum1 < m1 )
            m1 = sum1 ;
        if(sum1 > 0 )
            sum1 = 0 ;
        min2[i] = m1 ;

        sum += a[i] ;
        if( sum > m )
            m = sum ;
        if(sum < 0 )
            sum = 0 ;
        max1[i] = m ;

    }
    return ;
}

/*求n-0之间的最小值*/
void find_Min(int n )
{
    long long int i , sum = 0 , m = inf , sum1 = 0 , m1 = -inf  ;
    for( i = n-1 ;i >=0 ; i --)
    {
        sum +=a[i] ;
        if(sum < m )
            m = sum ;
        if(sum > 0 )
            sum = 0 ;
        min1[i] = m ;

        sum1 += a[i] ;
        if(sum1 > m1 )
            m1 = sum1;
        if(sum1 < 0 )
            sum1 = 0 ;
        max2[i] = m1 ;


    }
    return ;
}

int main()
{
    int t ;
    scanf("%d",&t);
    while(t--){
        int n ;
        scanf("%d",&n);
        int i ;
        for( i = 0 ; i < n ; i ++)
            cin >> a[i] ;
        memset(max1 , 0 , sizeof(max1)) ;
        memset(min1 , 0 , sizeof(min1)) ;
        memset(max2 , 0 , sizeof(max2)) ;
        memset(min2 , 0 , sizeof(min2)) ;
        find_Max(n);
        find_Min(n);

        long long int sum = 0 ;
        for( i = 0 ; i < n-1 ; i ++){
            if(sum < abs(max1[i] - min1[i + 1]))
                sum = abs(max1[i] - min1[i + 1]);
            if(sum < abs(max2[i + 1] - min2[i ]))
                sum = abs(max2[i + 1] - min2[i ]);
           // cout<<max1[i]<<" "<<min1[i + 1] <<"    "<<max2[i + 1]<<" "<<min2[i] << endl;
        }
        printf("%lld\n",sum);
    }
    return 0;
}