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
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.util.*;
import java.io.*;
/**
 *
 * @author billy
 */
public class MaxSumOnEvenPos2 {
    static class Pair implements Comparable<Pair>{
        int x;
        int y;

        Pair (int x0, int y0){
            x=x0;
            y=y0;

        }
        public int compareTo(Pair e){
            if (y>e.y){
                return 1;
            }
            else if (y<e.y){
                return -1;
            }
            else return 0;
        }
    }
        static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;



    public static void main(String[] args) throws IOException { 
       
        for (int t = readInt(); t>0; t--){
            int n = readInt();
            int[] a = new int[n];
            long og = 0;
            for (int i =0; i<n; i++){
                a[i] = readInt();
                if (i%2==0){
                    og+=a[i];
                }
            }
            long cs = 0;
            long mse = 0;
            
            for (int i =0; i<n-1; i+=2){
                cs = Math.max(cs+a[i+1]-a[i],0);
                mse = Math.max(cs,mse);
            }
            cs = 0;
            long mso = 0;
            
            for (int i =1 ;i<n-1; i+=2){
                cs = Math.max(cs+a[i]-a[i+1],0);
                mso = Math.max(cs,mso);
            }
            
            
            System.out.println(og + Math.max(mso,Math.max(mse,0)));
            
            
        }
        
    }





    static String read () throws IOException {
        while (st == null || !st.hasMoreTokens())
           st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    //© Maillew DMOJ
    static long readLong () throws IOException {
        return Long.parseLong(read());
    }
    static int readInt () throws IOException {
        return Integer.parseInt(read());
    }
    static double readDouble () throws IOException {
        return Double.parseDouble(read());
    }
    static char readChar () throws IOException {
        return read().charAt(0);
    }
    static String readLine () throws IOException {
        return br.readLine().trim();
    }
}
