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.math.BigInteger;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class D {

    public void solve() {
        final Scanner scanner = new Scanner(System.in);
        final int N = scanner.nextInt();
        for (int i = 0; i < N; i++) {
            final int n = scanner.nextInt();
            final int[] arr = new int[n];
            BigInteger e = BigInteger.valueOf(0);
            for (int j = 0; j < n; j++) {
                arr[j] = scanner.nextInt();
                if (j % 2 == 0) {
                    e = e.add(BigInteger.valueOf(arr[j]));
                }
            }
            BigInteger max = BigInteger.valueOf(0);
            Queue<Integer> queue = new LinkedList<>();
            boolean[] vis = new boolean[n];
            queue.add(0);
            queue.add(1);
            while (!queue.isEmpty()) {
                BigInteger prev = BigInteger.valueOf(0);
                int start = queue.poll();
                if (start > n - 1 || vis[start])
                    continue;
                vis[start] = true;
                for (int j = start; j < n - 1; j += 2) {
                    if (j % 2 == 0) {
                        prev = prev.add(BigInteger.valueOf(arr[j + 1] - arr[j]));
                    } else {
                        prev = prev.add(BigInteger.valueOf(arr[j] - arr[j + 1]));
                    }
                    if (prev.compareTo(max) > 0) {
                        max = new BigInteger(prev.toString());
                    } else if (prev.compareTo(BigInteger.valueOf(0)) <= 0) {
                        queue.add(j + 2);
                        break;
                    }
                }
            }
            System.out.println(max.add(e).toString());
        }
    }

    public static void main(final String[] args) {
        new D().solve();
    }
}