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.util.*;
import java.io.*;

public class MaxSumEvenPos {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(f.readLine());
        while(t-->0){
            int n = Integer.parseInt(f.readLine());
            long[] arr = new long[n];
            StringTokenizer st = new StringTokenizer(f.readLine());
            long currSum = 0;
            for(int i = 0; i < n; i++){
                arr[i] = Integer.parseInt(st.nextToken());
                if(i%2 == 0) currSum+=arr[i];
            }
            ArrayList<Long> arr1 = new ArrayList<Long>();
            ArrayList<Long> arr2 = new ArrayList<Long>();
            for(int i = 0; i+1 < n; i+=2){
                arr1.add(arr[i+1]-arr[i]);
            }
            for(int i = 1; i+1 < n; i+=2){
                arr2.add(arr[i]-arr[i+1]);
            }
            out.println(currSum + Math.max(mss(arr1), mss(arr2)));
        }
        out.close();
    }

    public static long mss(ArrayList<Long> arr){
        long max = 0;
        long currmax = 0;
        for(int i = 0; i < arr.size(); i++){
            currmax+=arr.get(i);
            if(max < currmax) max = currmax;
            if(currmax < 0) currmax = 0;
        }
        return max;

    }
}
