You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).

You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.

Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.

Example

Input


4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1


Output


26
5
37
5
import java.util.*;

public class codeforces
{

    private static long kadane(int arr[])
    {
        int start=0,end=0,n=arr.length;
        long sum=0,max=0;
        while(end<n)
        {
            sum+=arr[end];
            max=Math.max(sum,max);
            if(sum<=0)
            {
                while(start<end)
                {
                    sum-=arr[start++];
                    max=Math.max(sum,max);
                }

                start=end+1;
                sum=0;
            }
            end++;
        }

        while(start<end)
        {
            sum-=arr[start++];
            max=Math.max(max,sum);
        }

        return max;
    }

    public static void main(String[] args) 
    {
        Scanner s=new Scanner(System.in);
        int t=s.nextInt();

        while(t-->0)
        {
            int n=s.nextInt();

        int arr[]=new int[n];
        
        long sum=0;
        for(int i=0;i<n;i++)
        {  
            arr[i]=s.nextInt();
            if(i%2==0)
                sum+=arr[i];
        }

        int odd[]=new int[n/2+1];
        int even[]=new int[n/2+1];

        int index1=0,index2=0;
        for(int i=0;i<n;i++)
        {
            if(i%2==0&&i+1<n)
                even[index1++]=arr[i+1]-arr[i];
            else if(i%2==1&&i+1<n)
                odd[index2++]=arr[i]-arr[i+1];
        }

        long val1=kadane(even);
        long val2=kadane(odd);

        long max=Math.max(val1,val2);
        if(max>0)
            sum+=max;

        System.out.println(sum);
        }
    }
}