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
// package codeforces.educational_90;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public final class D {

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int t = Integer.parseInt(in.nextLine());
        for (int x = 0; x < t; x++) {
            final int n = Integer.parseInt(in.nextLine());
            final int[] arr = new int[n];
            long evenSum = 0;
            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
                if (i % 2 == 0) {
                    evenSum += arr[i];
                }
            }
            final long[] prefixSum = new long[n + 1];
            for (int i = 0; i < n; i++) {
                prefixSum[i + 1] = prefixSum[i] + (i % 2 == 0 ? -1 : 1) * arr[i];
            }
            final long[] minSums = { (long) 1e18, (long) 1e18 };
            long best = 0;
            for (int i = 0; i <= n; i++) {
                best = Math.max(best, prefixSum[i] - minSums[i % 2]);
                minSums[i % 2] = Math.min(minSums[i % 2], prefixSum[i]);
            }
            in.nextLine();
            System.out.println(evenSum + best);
        }
    }
}
