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.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.Collections;
import java.util.StringTokenizer;
 
public class Codeforces1373D {
 
	public static void main(String[] args) {
		FastScanner sc=new FastScanner();
		   StringBuilder sb=new StringBuilder();
	        int t=sc.nextInt();
	        while(t-->0){
	            int n=sc.nextInt();
	            int[] a=new int[n];
	            for(int i=0;i<n;i++)a[i]=sc.nextInt();
	            LinkedList<Integer> ll=new LinkedList<>();
	            for(int i=0;i<n;i++){
	                if(i%2==0 && i+1<n){
	                    ll.add(a[i+1]-a[i]);
	                }
	            }
	            long max=kadane(ll);
	            ll=new LinkedList<>();
	            for(int i=0;i<n;i++){
	                if(i%2==1 && i+1<n){
	                    ll.add(a[i]-a[i+1]);
	                }
	            }
	            max=Math.max(max,kadane(ll));
	            if(max<0)max=0;
	            for(int i=0;i<n;i++){
	                if(i%2==0)max+=a[i];
	            }
	            sb.append(max+"\n");
	        }
	        System.out.println(sb);
	    }
	    static long kadane(LinkedList<Integer> ll){
	        long curr_max=Long.MIN_VALUE,global_max=-1;
	        for(int i:ll){
	            if(curr_max==Long.MIN_VALUE){
	                curr_max=i;
	                global_max=i;
	            }
	            else{
	                curr_max=Math.max(curr_max+i,i);
	                if(curr_max>global_max)global_max=curr_max;
	            }
	        }
	        return global_max;
		
		
	}
	
	public void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		public String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}
 
}
