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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		String[] cmd=br.readLine().split(" ");
		int cases=Integer.valueOf(cmd[0]);
		while(cases!=0)
		{
			cases--;
			int n=Integer.valueOf(br.readLine());
			long[]arr=new long[n];
			cmd=br.readLine().split(" ");
			for(int i=0;i<n;i++)
			{
				arr[i]=Long.valueOf(cmd[i]);
			}
			ArrayList<Long> diff1=new ArrayList<>();
			ArrayList<Long> diff2=new ArrayList<>();
			long sum=0;
			for(int i=0;i<n;i=i+2)
			{
				if(i+1<n)
				diff1.add(arr[i+1]-arr[i]);
				sum=sum+arr[i];
			}
			for(int i=1;i<n;i=i+2)
			{
				if(i+1<n)
				diff2.add(arr[i]-arr[i+1]);
			}
			long curr=0;
			long mx=0;
			for(int i=0;i<diff1.size();i++)
			{
				curr=curr+diff1.get(i);
				mx=Math.max(mx, curr);
				if(curr<0)
					curr=0;
			}
			curr=0;
			for(int i=0;i<diff2.size();i++)
			{
				curr=curr+diff2.get(i);
				mx=Math.max(mx,curr);
				if(curr<0)
					curr=0;
			}
			sum=sum+mx;
			System.out.println(sum);
		}
	}

}
