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
//package CodeForces;

import java.util.ArrayList;
import java.util.Scanner;

public class MaximumSumOnEvenPosition {
	
	static long solve(ArrayList<Long> arr)
	{
		long max=Integer.MIN_VALUE;
		long sum=0;
		for(int i=0;i<arr.size();i++)
		{
			sum=sum+arr.get(i);
			max=Math.max(max,sum);
			if(sum<0)
			{
				sum=0;
			}
		}
		return max;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		while(t>0)
		{
			int n=s.nextInt();
			long arr[]=new long[n];
			long e=0;long o=0;
			for(int i=0;i<n;i++)
			{
				arr[i]=s.nextLong();
				if(i%2==0)
				{
					e=e+arr[i];
				}
				else
				{
					o=o+arr[i];
				}
			}
			ArrayList<Long> a=new ArrayList<Long>();
			ArrayList<Long> b=new ArrayList<Long>();
			for(int i=0;i+1<arr.length;i=i+2)
			{
				a.add(arr[i+1]-arr[i]);
			}
			for(int i=1;i+1<arr.length;i=i+2)
			{
				b.add(arr[i]-arr[i+1]);
			}
			long x=solve(a);
			long y=solve(b);
			System.out.println(e+Math.max(0,Math.max(x,y)));
			t--;
		}

	}

}
