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.*;
import java.io.*;
import java.math.*;
public final class CF
{
    static int K=(int)Math.pow(10,9)+7;
     public static void main(String[]args)throws IOException
	{
        FastReader ob=new FastReader();
        StringBuffer sb=new StringBuffer();
        int T=ob.nextInt();
        while(T-->0)
        {
            int n=ob.nextInt();
            int a[]=new int[n];
            long sum=0;
            for(int i=0;i<n;i++)
            {
                a[i]=ob.nextInt();
                if(i%2==0)
                sum+=a[i];
            }
            ArrayList<Integer> b=new ArrayList<>();
            ArrayList<Integer> c=new ArrayList<>();
            long maxsum1=0,sum1=0;int k=0;
            for(int i=1;i<n;i+=2)
            {
                b.add(a[i]-a[i-1]);
                sum1+=b.get(k);
                maxsum1=Math.max(maxsum1, sum1);
                if(sum1<0)
                sum1=0;
                ++k;
            }
            k=0;
            //System.out.println(b);
            long maxsum2=0,sum2=0;
            for(int i=1;i<n-1;i+=2)
            {
                c.add(a[i]-a[i+1]);
                sum2+=c.get(k);
                maxsum2=Math.max(maxsum2, sum2);
                if(sum2<0)
                sum2=0;
                ++k;
            }
            //System.out.println(c);
            sum+=Math.max(maxsum1,maxsum2);
            sb.append(sum+"\n");
        }
        System.out.println(sb);
	}
}
class Pair
{
	int x,y;
	Pair(int x,int y)
	{
		this.x=x;
		this.y=y;
	}
}
class PairComparator implements Comparator<Pair>
{
	public int compare(Pair a,Pair b)
	{
		return a.x-b.x;
	}
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;
    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e)  {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    public String nextLine()
    {
        String s="";
        try {
        s=br.readLine();
        } catch (IOException e)  {
                e.printStackTrace();
            }
        return s;
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public double nextDouble() {
        return Double.parseDouble(next());
    }
}