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
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300IQ as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class D
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         while(T-->0)
         {
            int N = Integer.parseInt(infile.readLine());
            int[] arr = new int[N];
            st = new StringTokenizer(infile.readLine());
            for(int i=0; i < N; i++)
               arr[i] = Integer.parseInt(st.nextToken());
            long res = arr[0];
            if(N >= 2)
            {
               for(int i=2; i < N; i+=2)
                  res += arr[i];
               ArrayList<Integer> ls = new ArrayList<Integer>();
               for(int i=1; i < N; i+=2)
                  ls.add(arr[i]-arr[i-1]);
               long val = kd(ls, 1);
               ls = new ArrayList<Integer>();
               for(int i=1; i+1 < N; i+=2)
                  ls.add(arr[i]-arr[i+1]);
               val = Math.max(val, kd(ls, 1));
               res += val;
            }
            sb.append(res+"\n");
         }
         System.out.print(sb);
      }
      public static long kd(ArrayList<Integer> ls, int mult)
      {
         long curr = 0L;
         long max = 0L;
         for(int x: ls)
         {
            curr += x*mult;
            max = Math.max(curr, max);
            curr = Math.max(curr, 0);
         }
         return max;
      }
   }