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.Scanner;

public class CodeForce1373D {
    public static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);
        int cases = Integer.parseInt(scanner.nextLine());
        while (cases-- > 0) {
            scanner.nextLine();
            solve(scanner.nextLine());
        }
    }

    private static void solve(String s) {
        String[] ip = s.split(" ");
        long[] conv = new long[ip.length];
        for (int i = 0; i < ip.length; i++) {
            conv[i] = Long.parseLong(ip[i]);
        }

        long evenSum = 0;

        for (int i = 0; i < conv.length; i++)
            if (i % 2 == 0) evenSum += conv[i];

        long maxDif=0, perSum = 0;
        for (int i = 1; i < ip.length; i+=2) {
            perSum = Math.max(perSum + conv[i] - conv[i - 1], 0);
            maxDif = Math.max(maxDif, perSum);

        }
        perSum = 0;
        for (int i = 1; i < ip.length-1; i+=2) {
            perSum = Math.max(perSum + conv[i] - conv[i + 1], 0);
            maxDif = Math.max(maxDif, perSum);

        }
        System.out.println(evenSum+ maxDif);
    }
}
