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.*;
import java.util.*;
public class Main
{
	
	private void solve()throws Exception
	{
		int n=nextInt();
		int a[]=new int[n];
		long sum[]=new long[n];
		long evensum=0;
		for(int i=0;i<n;i++)
		{
			a[i]=nextInt();
			if(i%2==0)
			{
				sum[i]=(i==0?0:sum[i-1])-a[i];
				evensum+=a[i];
			}
			else
				sum[i]=sum[i-1]+a[i];
		}
		long maxodd=-Long.MAX_VALUE;
		long maxeven=-Long.MAX_VALUE;
		boolean mo=false,me=false;
		long ans=evensum;
		for(int i=n-1;i>=0;i--)
		{
			if(i%2==0)
			{
				if(mo)
					ans=Math.max(ans,evensum+maxodd-(i==0?0:sum[i-1]));
				maxeven=Math.max(maxeven,sum[i]);
				me=true;
			}
			else
			{
				if(me)
					ans=Math.max(ans,evensum+maxeven-sum[i-1]);
				maxodd=Math.max(maxodd,sum[i]);
				mo=true;
			}
		}
		out.println(ans);
	}

	 
	///////////////////////////////////////////////////////////

	public void run()throws Exception
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		int t=nextInt();
		while(t-->0)
		solve();
		
		br.close();
		out.close();
	}
	public static void main(String args[])throws Exception{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws Exception{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws Exception{
		return br.readLine();
	}
	int nextInt()throws Exception{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws Exception{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws Exception{
		return Double.parseDouble(nextToken());
	}
}