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.*;
import java.lang.*;
 
 
public class Main{
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(new BufferedOutputStream(System.out));
        int T = in.nextInt();
        for(int test=0;test<T;test++){
            int n = in.nextInt();
            long[] a = new long[n];
            for(int i=0;i<n;i++){
                a[i]=in.nextInt();
            }
            if(n==1){
                out.println(a[0]);
                continue;
            }
            long[] dp = new long[n];
            for(int i=1;i<n;i++){
                long sub = (i-2)>0 ? dp[i-2] : 0;
                if(i%2==0){
                    dp[i]-=a[i];
                    dp[i]+=a[i-1];
                } else {
                    dp[i]+=a[i];
                    dp[i]-=a[i-1];
                }
                dp[i]+=Math.max(sub,0);
            }
            long mx=dp[1];
            for(int i=1;i<n;i++){
                mx=Math.max(mx,dp[i]);
            }
            long ans = 0;
            for(int i=0;i<n;i++){
                if(i%2==0){
                    ans+=a[i];
                }
            }
            ans+=Math.max(mx,0);
            out.println(ans);
            
        }

        out.close();
   }

     

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;
}