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.Arrays;
import java.util.Scanner;
public class edu90
{
	public static void main (String[] args)
	 {
		Scanner sc=new Scanner(System.in); 
		int t,i;
		t=sc.nextInt();
		while(t>0)
		{
			t--;
			int n=sc.nextInt();
			long a[]=new long[n];
			for(i=0;i<n;i++)
				a[i]=sc.nextInt();
			if(n==1)
			{
				System.out.println(a[0]);
				continue;
			}
			long ans=0;
			for(i=0;i<n;i+=2)
				ans+=a[i];
			
			long k1[]=new long[n/2+1];
			long k2[]=new long[n/2+1];
			k1[0]=a[1]-a[0];
			long max=0;
			if(k1[0]>max)
				max=k1[0];
			int j=1;
			
			for(i=3;i<n;i+=2)
			{
				long v=a[i]-a[i-1];
				k1[j]=Math.max(0, k1[j-1]+v);
				k1[j]=Math.max(k1[j], v);
				if(k1[j]>max)
					max=k1[j];
				j++;
			}
			if(n>2)
				k2[0]=a[1]-a[2];
			j=1;
			if(k2[0]>max)
				max=k2[0];
			for(i=3;i<n;i+=2)
			{
				long v=0;
				if(i+1<n)
					v=a[i]-a[i+1];
				k2[j]=Math.max(0, k2[j-1]+v);
				k2[j]=Math.max(k2[j], v);
				if(k2[j]>max)
					max=k2[j];
				j++;
			}
			System.out.println(max+ans);
		}
	}
}
