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 maxSum {
    public static void main(String[] args) throws Exception{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(in.readLine());
        for(int i = 0; i < t; i++){
            int n = Integer.parseInt(in.readLine());
            int[] arr = new int[n];
            StringTokenizer st = new StringTokenizer(in.readLine());
            long sum = 0;
            for(int j = 0; j < n; j++){
                arr[j] = Integer.parseInt(st.nextToken());
                if(j % 2 == 0){
                    sum += arr[j];
                }
            }
            //even
            int end = ((n - 1) % 2 == 0) ? n - 1 : n - 2;
            long LM = Long.MIN_VALUE + Integer.MAX_VALUE, GM = Long.MIN_VALUE + Integer.MAX_VALUE;
            for(int j = 2; j <= end; j += 2){
                int A = arr[j - 1] - arr[j];
                LM = Math.max(A, A + LM);
                if(LM > GM){
                    GM = LM;
                }
            }
            LM = Long.MIN_VALUE + Integer.MAX_VALUE;
            end = ((n - 1) % 2 == 1) ? n - 1 : n - 2;
            for(int j = 1; j <= end; j += 2){
                int A = arr[j] - arr[j - 1];
                LM = Math.max(A, A + LM);
                if(LM > GM){
                    GM = LM;
                }
            }
            if(GM > 0){
                System.out.println(sum + GM);
            }
            else{
                System.out.println(sum);
            }
        }
        in.close();
    }
}
