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.ArrayList;
import java.util.Scanner;

public class Main {

    private static long maxSum(ArrayList<Integer> a) {
        long ans = (long) -(1e15 + 7);
        long curr = 0;
        for (int i = 0; i < a.size(); i++) {
            if (a.get(i) >= curr + a.get(i)) {
                curr = a.get(i);
            } else curr += a.get(i);
            ans = Math.max(ans, curr);
        }
        return ans;
    }

    public static void main(String[] args) {
        int t = 1;
        Scanner sc = new Scanner(System.in);
        t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int a[] = new int[n];
            long sum = 0;
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                if ((i & 1) == 0) {
                    sum += a[i];
                }
            }
            ArrayList<Integer> even = new ArrayList<>();
            ArrayList<Integer> odd = new ArrayList<>();
            for (int i = 0; i < n - 1; i += 2) {
                even.add(a[i + 1] - a[i]);
            }
            for (int i = 1; i < n - 1; i += 2) {
                odd.add(a[i] - a[i + 1]);
            }
            sum += Math.max(0, Math.max(maxSum(even), maxSum(odd)));
            System.out.println(sum);
        }
    }
}
