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 er90;

import java.util.Scanner;

public class fourth {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0)
		{
			int n = sc.nextInt();
			int arr[] = new int[n];
			long evensum = 0;
			for(int i =0;i<n;i++)
			{
				arr[i] = sc.nextInt();
				if(i%2==0)
					evensum+=arr[i];
			}
			int sum[] = new int[n];
			
			for(int i =0;i<n;i++)
			{
				if(i%2==0 && i<n-1)
				{
					sum[i] = arr[i+1]-arr[i];
					
				}
				else if(i<n-1)
				{
					sum[i] = arr[i] - arr[i+1];
				}
				//System.out.println(sum[i]);
			}
			//System.out.println(evensum);
			int size = sum.length; 
	        long max_so_far = Long.MIN_VALUE, max_ending_here = 0; 
	  
	        for (int i = 0; i < size; i+=2) 
	        { 
	            max_ending_here = max_ending_here + sum[i]; 
	            if (max_so_far < max_ending_here) 
	                max_so_far = max_ending_here; 
	            if (max_ending_here < 0) 
	                max_ending_here = 0; 
	        } 
	        long max_so_far2 = Long.MIN_VALUE, max_ending_here2 = 0; 
	  	  
	        for (int i = 1; i < size; i+=2) 
	        { 
	            max_ending_here2 = max_ending_here2 + sum[i]; 
	            if (max_so_far2 < max_ending_here2) 
	                max_so_far2 = max_ending_here2; 
	            if (max_ending_here2 < 0) 
	                max_ending_here2 = 0; 
	        } 
	        
	        long max = Math.max(0,Math.max(max_so_far2, max_so_far));
	        System.out.println(max+evensum);
			
			
		}

	}

}
