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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class ProblemD {
    public static void main(String[] args) throws Exception {
        Scanner scanner = new Scanner(System.in);
        int tot = scanner.nextInt();
        while (tot-- > 0) {
            long n = scanner.nextLong();
            List<Long> nums = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                nums.add(scanner.nextLong());
            }
            List<Long> rev_nums = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                rev_nums.add(nums.get((int)n-i-1));
            }

            // 짝수개면 한번만
            if(n%2 == 0){
                System.out.println(Math.max(get_3(nums), get_2(nums)));
            } else {
                System.out.println(Math.max(get(nums), get(rev_nums)));
            }

        }
    }
    public static long get_3(List<Long> nums){
        int n = nums.size();
        List<Long> nums2 = new ArrayList<>();
        long ret = 0;
        for (int i = 0; i < n; i++) {
            if(i%2 == 0)
                ret += nums.get(i);
        }
        for (int i = 0; i < n/2-1; i++) {
            nums2.add(nums.get(2*i+1) - nums.get(2*i+2));
        }
        long tempret = 0;
        long max = 0;
        for (int i = 0; i < nums2.size(); i++) {
            tempret += nums2.get(i);
            if(tempret < 0)
                tempret = 0;
            max = Math.max(tempret, max);
        }

        return (ret+max);
    }
    public static long get_2(List<Long> nums){
        int n = nums.size();
        List<Long> nums2 = new ArrayList<>();
        long ret = 0;
        for (int i = 0; i < n/2; i++) {
            nums2.add(nums.get(2*i+1) - nums.get(2*i));
            ret+=nums.get(2*i);
        }
        long tempret = 0;
        long max = 0;
        for (int i = 0; i < nums2.size(); i++) {
            tempret += nums2.get(i);
            if(tempret < 0)
                tempret = 0;
            max = Math.max(tempret, max);
        }

        return (ret+max);
    }
    public static long get(List<Long> nums){
        int n = nums.size();
        List<Long> nums2 = new ArrayList<>();
        long ret = nums.get(nums.size()-1);
        for (int i = 0; i < n/2; i++) {
            nums2.add(nums.get(2*i+1) - nums.get(2*i));
            ret+=nums.get(2*i);
        }
        long tempret = 0;
        long max = 0;
        for (int i = 0; i < nums2.size(); i++) {
            tempret += nums2.get(i);
            if(tempret < 0)
                tempret = 0;
            max = Math.max(tempret, max);
        }

        return (ret+max);
    }
}
