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.awt.*;
import java.util.*;
import java.io.*;
import java.math.BigDecimal;
import java.nio.file.Files;
import java.nio.charset.StandardCharsets;
//给出两个数a,b 求出[a,b]中各位数字之和能整除原数的数的个数。
public class Main {

    // 5
    // oof

    int mod  = (int) 1e9+7;
    //
    //int n = (int) 1e6+5;
    //long[][] memo = new long[n][27];
    void solve(int n, int[] arr) {
        PriorityQueue<Long> pq1 = new PriorityQueue<>();//store odd
        PriorityQueue<Long> pq2 = new PriorityQueue<>();//store even
        long e = 0;
        long o = 0;
        long res = 0;
        for(int i=0;i<n;i++){
            if(i%2==0) res+=arr[i];
        }
        pq1.add((long)0);
        pq2.add((long)0);
        long max = res;
        //System.out.println("res:"+res);
        for(int i=0;i<n;i++){
            long dis = -1;
            if(i%2==0){
                e+=arr[i];
                long tot = o-e;
                if(!pq2.isEmpty()){
                    long v = pq2.peek();
                    dis = tot-v;
                    max = Math.max(max,res+dis);
                }
                //System.out.println("even:"+" o:"+o+" e:"+e+" "+pq1+" "+pq2+" "+dis);
                pq2.add(tot);

            }else{
                o+=arr[i];
                long tot = o-e;
                if(!pq1.isEmpty()){
                    long v = pq1.peek();
                    dis = tot-v;
                    max = Math.max(max,res+dis);
                }
                //System.out.println("odd:"+" o:"+o+" e:"+e+" "+pq1+" "+pq2+" "+dis);
                pq1.add(tot);
            }
        }
        System.out.println(max);
    }
    //0 1 2 3 4 5 6 7
    //1 7 3 4 7 6 2 9
    //1 7 4 3 7 6 2 9;
    //1 2 1 2 1
    public static void main(String[] args)  throws IOException {
        // Use the Scanner class

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String[]  s = br.readLine().split(" ");
        Main test = new Main();

        int n = Integer.parseInt(s[0]);
        //s = br.readLine().split(" ");
        //s = br.readLine().split(" ");
        for(int i=0;i<n;i++){
            s = br.readLine().split(" ");
            int len = Integer.parseInt(s[0]);
            s = br.readLine().split(" ");
            int[] arr = new int[len];
            for(int j=0;j<len;j++){
                arr[j] = Integer.parseInt(s[j]);
            }
            test.solve(len,arr);

        }

    }

}
