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.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException{
        //BufferedReader f = new BufferedReader(new FileReader("uva.in"));
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(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());
            int[] a = new int[n];
            for(int j = 0; j < n; j++) {
                a[j] = Integer.parseInt(st.nextToken());
            }
            long even = 0;
            for(int j = 0; j < n; j += 2) {
                even += a[j];
            }
            long option1 = 0;
            long option2 = 0;
            long maxBenefit = 0;
            for(int j = 0; j < n; j += 2) {
                if(j > 0) {
                    option1 += a[j-1]-a[j];
                    maxBenefit = Math.max(maxBenefit, option1);
                    if(option1 < 0) {
                        option1 = 0;
                    }
                }
                if(j < n-1) {
                    option2 += a[j+1]-a[j];
                    maxBenefit = Math.max(maxBenefit, option2);
                    if(option2 < 0) {
                        option2 = 0;
                    }
                }
            }
            out.println(even+maxBenefit);
        }
        f.close();
        out.close();
    }
}