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.InputStreamReader;

public class MaxEven {

    public static void main(String[] args) {

        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            long t = Long.valueOf(br.readLine());
            for (long t_ = 0; t_ < t; t_++) {
                int length = Integer.valueOf(br.readLine());
                String[] a = br.readLine().split(" ");
                System.out.println(solve(length, a));
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    private static long solve(int length, String[] a_) {
        long[] a = new long[length];
        for (int i = 0; i < length; i++) {
            a[i] = Integer.valueOf(a_[i]);
        }
        long maxAddition = 0;
        long maxInRange = 0;
        for (int i = 0; i < length-1; i+= 2) {
            maxInRange += a[i+1] - a[i];
            maxInRange = Math.max(0, maxInRange);
            maxAddition = Math.max(maxAddition, maxInRange);
        }
        maxInRange = 0;
        // not changing the 1st
        for (int i = 1; i < length-1; i+=2) {
            maxInRange += a[i] - a[i+1];
            maxInRange = Math.max(0, maxInRange);
            maxAddition = Math.max(maxAddition, maxInRange);
        }
        long total = 0;
        for (int i = 0; i < length; i+=2) {
            total += a[i];
        }
        total += maxAddition;
        return total;
    }
}
