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 solve{
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            long[] ar=new long[n];
            
            long[] od=new long[n];
            long[] ev=new long[n];
            long ans=0;
            
            for(int i=0;i<n;i++) {ar[i]=sc.nextLong(); if(i%2==0) ans+=ar[i];}
            
            for(int i=0;i<n;i+=2){
                if(i+1<n){
                ev[i]=(ar[i+1])-(ar[i]);
               
                    
                }
            }
            
             for(int i=1;i<n;i+=2){
                 if(i+1<n){
                od[i]=(ar[i])-(ar[i+1]);
                
              
                 }
            }
    
            if(Math.max(maxsum(od),maxsum(ev))>0){
                ans+=Math.max(maxsum(od),maxsum(ev));
            }
            
            System.out.println(ans);
            
        }
        
    }
    static long maxsum(long[] ar){
        long an=0;
        long sum=0;
        for(int i=0;i<ar.length;i++){
             sum+=(ar[i]);
          
            
            if(sum>0){
             
            
            an=Math.max(an,sum);
            
            }
            else if(sum<0){
                sum=0;
            }
            
        }
        return an;
    }
}