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
//package test.codeforce.r90;

import java.util.Scanner;

/**
 * @author Lin
 * @version 1.0
 * @date 2020/6/25 23:48
 */
public class D
{

	public static void main(String[] args)
	{
		Scanner input = new Scanner(System.in);
		int testNum = input.nextInt();
		StringBuffer stringBuffer = new StringBuffer();
		for (int rr = 0; rr < testNum; rr++)
		{
			int n = input.nextInt();

			long ans = 0;
			//������������
			int[] an = new int[n];
			for (int i = 0; i < n; i++)
			{
				an[i] = input.nextInt();
				if (i % 2 == 0)
				{
					ans += an[i];
				}
			}
			if (n == 1)
			{
				ans = an[0];
			} else if (n == 2)
			{
				ans = Math.max(an[0], an[1]);
			}
			else
			{
				int[] s = new int[n / 2];
				int[] t = new int[(n - 1) / 2];
				for (int i = 0; i < n - 1; i = i + 2)
				{
					s[i / 2] = an[i + 1] - an[i];
				}
				for (int i = 1; i < n - 1; i = i + 2)
				{
					t[i / 2] = an[i] - an[i + 1];
				}
				long x, y;
				x = Math.max(s[0], 0);
				y = Math.max(t[0], 0);
				long ss = x;
				long tt = y;
				for (int i = 1; i < n / 2; i++)
				{
					x = Math.max(x + s[i], s[i]);

					ss = Math.max(ss, x);
				}
				for (int i = 1; i < (n - 1) / 2; i++)
				{
					y = Math.max(y + t[i], t[i]);

					tt = Math.max(tt, y);
				}
				ans = ans + Math.max(tt, ss);
			}

			stringBuffer.append(ans + "\n");
		}
		System.out.println(stringBuffer.toString());
	}
}
