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.*;
public class scratch_18{


    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine());
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        static long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }


    public static void main(String[] args)  throws IOException {
        Reader.init(System.in);
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
        int t= Reader.nextInt();
        for (int tt = 0; tt <t ; tt++) {
            int n= Reader.nextInt();
            long arr[]= new long[n];
            for (int i = 0; i <n ; i++) {
                arr[i]= Reader.nextLong();
            }
//            for (int i = 0; i <n ; i++) {
//                System.out.print(i+" ");
//            }
//            System.out.println();
            long dp[][]= new long[n+2][3];  // 0 if not started, 1 if going on, 2 if ended. (reversal)

            // this is the loop for even starting postion subarrays.
            for (int i = 0; i <n ; i++) {
                dp[i+1][0]= Math.max(dp[i+1][0],dp[i][0]+((i%2==0)?arr[i]:0));  // not started
                dp[i+1][1]= Long.MIN_VALUE;                               // reversal has started. the previous might or might not be in reversal
                if(i%2==0){
                    dp[i+1][1]= Math.max(dp[i+1][1],Math.max(dp[i][1],dp[i][0]));   // since the starting position needs to be even , so the previous one could not have been in it
                }
                if(i%2==1){
                    dp[i+1][1]= Math.max(dp[i+1][1],dp[i][1]+arr[i]);    // the previous one might be it or might not be it(the starting position)
                }
                dp[i+1][2]= Long.MIN_VALUE;
                if(i%2==1){
                    dp[i+1][2]= Math.max(dp[i+1][2],dp[i][2]);
                    // either ended on pichla or ended before pichla.
                }

                if(i%2==0){
                    dp[i+1][2]= Math.max(dp[i+1][2],Math.max(dp[i][2],dp[i][1])+arr[i]);
                }
                // we still need to ensure that the subarray is of even length
            }
//



            if(n%2==0){
                dp[n+1][2]= Math.max(dp[n+1][2],Math.max(dp[n][2],dp[n][1]));
            }
            else{
                dp[n+1][2]= Math.max(dp[n+1][2],dp[n][2]);
            }



//            for (int i = 0; i <=n+1 ; i++) {
//                System.out.println(Arrays.toString(dp[i]));
//            }


//            System.out.println();

            long max= Math.max(0,dp[n+1][2]);

            dp= new long[n+2][3];


                // start with odd
            for (int i = 0; i <n ; i++) {

                dp[i+1][0]= Math.max(dp[i+1][0],dp[i][0]+((i%2==0)?arr[i]:0));  // not started

                dp[i+1][1]= Long.MIN_VALUE;                               // reversal has started. the previous might or might not be in reversal

                if(i%2==0){
                    dp[i+1][1]= Math.max(dp[i+1][1],dp[i][1]);
                }
                if(i%2==1){
                    dp[i+1][1]= Math.max(dp[i+1][1],Math.max(dp[i][1],dp[i][0])+arr[i]);   // the previous one might be it or might not be it(the starting position)
                }


                dp[i+1][2]= Long.MIN_VALUE;

                if(i%2==0){
                    dp[i+1][2]= Math.max(dp[i+1][2],dp[i][2]+arr[i]);
                }

                if(i%2==1){
                    dp[i+1][2]= Math.max(dp[i+1][2],Math.max(dp[i][1],dp[i][2])) ;
                }
            }
//            System.out.println();

//            for (int i = 0; i <=n ; i++) {
//                System.out.println(Arrays.toString(dp[i]));
//            }

            if(n%2==0){
                dp[n+1][2]= Math.max(dp[n+1][2],dp[n][2]);
            }
            else{
                dp[n+1][2]= Math.max(dp[n+1][2],Math.max(dp[n][2],dp[n][1]));
            }
            max= Math.max(max,dp[n+1][2]);
            out.append(max+"\n");
        }
        out.flush();
        out.close();
    }
}








