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.*;
import java.io.*;
 
public class Solution{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] input=br.readLine().trim().split(" ");
		int numTestCases=Integer.parseInt(input[0]);
		while(numTestCases-->0) {
			input=br.readLine().trim().split(" ");
			int n=Integer.parseInt(input[0]);
			int[] arr=new int[n];
			input=br.readLine().trim().split(" ");
			for(int i=0;i<n;i++) {
				arr[i]=Integer.parseInt(input[i]);
			}
			out.println(maxSum(arr));
		}
		out.flush();
		out.close();
	}		
	public static long maxSum(int[] arr)
	{
		int n=arr.length;
		if(n==1) {
			return arr[0];
		}
		long maxSum=0;
		long currSum=0;
		for(int i=0;i<n;i+=2)
		{
			maxSum+=arr[i];
			currSum+=arr[i];
		}
		int[] temp=new int[n/2];
		int k=0;
		for(int i=0;i<n;i+=2) {
			if(i+1<n) {
				temp[k]=arr[i+1]-arr[i];
			}
			k++;
		}
		
		
		long sum=currSum+kadane(temp);
		
		maxSum=Math.max(maxSum, sum);
		
		k=0;
		temp=new int[n/2];
		for(int i=1;i<n;i+=2) {
			if(i+1<n) {
				temp[k]=arr[i]-arr[i+1];
			}
			k++;
		}
		
		
		sum=currSum+kadane(temp);
		
		maxSum=Math.max(maxSum, sum);
		
		return maxSum;
	}
	
	public static long kadane(int[] arr)
	{
		int n=arr.length;
		long maxSum=arr[n-1];
		long sum=arr[n-1];
		for(int i=n-2;i>=0;i--) {
			sum=Math.max(arr[i], sum+arr[i]);
			maxSum=Math.max(maxSum, sum);
		}
		return maxSum;
	}
}