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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
//import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.StringTokenizer;

public class MaximumSumOnEvenPositions { // Template for CF
    public static class ListComparator implements Comparator<List<Integer>> {

        @Override
        public int compare(List<Integer> l1, List<Integer> l2) {
            for (int i = 0; i < l1.size(); ++i) {
                if (l1.get(i).compareTo(l2.get(i)) != 0) {
                    return l1.get(i).compareTo(l2.get(i));
                }
            }
            return 0;
        }
    }

    public static class Pair {
        int first;
        int second;

        public Pair(int first, int second) {
            this.first = first;
            this.second = second;
        }

        public int getFirst() {
            return first;
        }

        public int getSecond() {
            return second;
        }

        @Override
        public String toString() {
            return first + " " + second;
        }
    }

    public static void main(String[] args) throws IOException {
        // Check for int overflow!!!!
        // Should you use a long to store the sum or smthn?
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int T = Integer.parseInt(f.readLine());
        for (int i = 0; i < T; i++) {
            int n = Integer.parseInt(f.readLine());
            StringTokenizer st = new StringTokenizer(f.readLine());
            List<Integer> list = new ArrayList<>();
            for (int j = 0; j < n; j++) {
                int a = Integer.parseInt(st.nextToken());
                list.add(a);
            }
            long[][] dp = new long[3][n + 1];
            for (int j = 0; j < n; j++) {
                dp[0][j + 1] = dp[0][j] + ((j % 2 == 0) ? list.get(j) : 0);
                if (j + 2 <= n) {
                    dp[1][j + 2] = Math.max(dp[0][j], dp[1][j]) + ((j % 2 == 0) ? list.get(j + 1) : list.get(j));
                }
                dp[2][j + 1] = Math.max(Math.max(dp[0][j], dp[1][j]), dp[2][j]) + ((j % 2 == 0) ? list.get(j) : 0);
            }

            /*
             * for (int j = 0; j < 3; j++) { System.out.println(Arrays.toString(dp[j])); }
             */
            out.println(Math.max(Math.max(dp[0][n], dp[1][n]), dp[2][n]));
        }
        out.close();
    }

}