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 D{
	private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	private static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
	
	public static void main (String[] args) throws IOException{
		int t = Integer.parseInt(br.readLine());
		while(t-- > 0){
            int n = Integer.parseInt(br.readLine());
            StringTokenizer st = new StringTokenizer(br.readLine());
            int a[] = new int[n];
            for(int i=0;i<n;i++){
                a[i] = Integer.parseInt(st.nextToken());
            }
            long even = 0, odd = 0;
            long max = 0;
            for(int i=0;i<n;i++){
                if(n%2!=0 && i==n-1){
                    break;
                }
                if(i%2==0){
                    even += a[i];
                }else{
                    odd += a[i];
                }
                if(i%2!=0){
                    long diff = odd - even;
                    if(diff>0){
                        if(diff>max){
                            max = diff;
                        }
                    }else{
                        odd = 0;
                        even = 0;
                    }
                }
            }
            even = 0;
            odd = 0;
            for(int i=1;i<n;i++){
                if(n%2==0 & i==n-1){
                    break;
                }
                if(i%2==0){
                    even += a[i];
                }else{
                    odd += a[i];
                }
                if(i%2==0){
                    long diff = odd - even;
                    if(diff>0){
                        if(diff>max){
                            max = diff;
                        }
                    }else{
                        odd = 0;
                        even = 0;
                    }
                }
            }
            long sum = max;
            for(int i=0;i<n;i++){
                if(i%2 == 0){
                    sum += a[i];
                }
            }
            bw.write(sum + "");
            bw.newLine();
		}
		bw.flush();
	}
}
