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.*;
public class Main{
    public static long find(int[] arr){
            long sum=0,cur=0;
            for(int i=0;i<arr.length;i++){
                if(cur<0) cur=arr[i];
                else{
                    cur=cur+arr[i];
                }
                sum=Math.max(sum,cur);
            }
            return sum;
    }
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int[] arr=new int[n];
            for(int i=0;i<n;i++) arr[i]=sc.nextInt();
            long ans=0;
            for(int i=0;i<n;i+=2){
                ans+=arr[i];
            }
            if(n%2==0){
                int[] arr1=new int[n/2];
                int[] arr2=new int[n/2-1];
                for(int i=0;i<n;i+=2){
                    arr1[i/2]=arr[i+1]-arr[i];
                }
                for(int i=1;i<n-1;i+=2) arr2[i/2]=arr[i]-arr[i+1];
                long sum1=find(arr1);
                long sum2=find(arr2);
                ans+=Math.max(sum2,sum1);
            }
            else{
                int[] arr1= new int[n/2];
                int[] arr2=new int[n/2];
                for(int i=0;i<n-1;i+=2) arr1[i/2]=arr[i+1]-arr[i];
                for(int i=1;i<n;i+=2) arr2[i/2]=arr[i]-arr[i+1];
                long sum1=find(arr1);
                long sum2=find(arr2);
                ans+=Math.max(sum1,sum2);
            }
            System.out.println(ans);
        }
    }
}