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
	// _________________________RATHOD_____________________________________________________________
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
public class Main 
{ 
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !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()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
  
    public static void main(String[] args) 
    { 
        FastReader sc=new FastReader(); 
        int t=sc.nextInt();
        while(t>0)
        {
        t--;    
        int n=sc.nextInt();
        int ar[]=new int[n];
        int i;
         long sum=0;
        for(i=0;i<n;i++)
        {
        ar[i]=sc.nextInt();
        if(i%2==0)
        sum=sum+((long) ar[i]);
        }
        
        ArrayList<Integer> sa1=new ArrayList<Integer>();
        ArrayList<Integer> sa2=new ArrayList<Integer>();
        for(i=1;i<(n);i=i+2)
        {
            sa1.add(ar[i]-ar[i-1]);
        }
        for(i=2;i<(n);i=i+2)
        {
            sa2.add(ar[i-1]-ar[i]);
        }
        
        //System.out.println(sa[n/2-1]);
        long max1=maxSubArraySum(sa1);
        long max2=maxSubArraySum(sa2);
        //System.out.println(max1+" "+max2);
        long max=Math.max(max1,max2);
       if(max<0)
System.out.println(sum);
       else
        System.out.println((max)+sum);
        }
}
 static long maxSubArraySum(ArrayList<Integer> a) 
    { 
        int size = a.size(); 
        long max_so_far = Long.MIN_VALUE, max_ending_here = 0; 
  
        for (int i = 0; i < size; i++) 
        { 
            max_ending_here = max_ending_here + ((long) a.get(i)); 
            if (max_so_far < max_ending_here) 
                max_so_far = max_ending_here; 
            if (max_ending_here < 0) 
                max_ending_here = 0; 
        } 
        return max_so_far; 
    } 
}