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 Solution
{
 
 
  static long mx(int a[]) 
    { 
        int size = a.length; 
        long max_so_far = Long.MIN_VALUE, max_ending_here = 0; 
  
        for (int i = 0; i < size; i++) 
        { 
            max_ending_here = max_ending_here + a[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; 
    } 
    
    
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int t = sc.nextInt();
              sc.nextLine();
              while(t-- > 0)
              {
                  int n = sc.nextInt();
                  int arr[] = new int[n];
                  long sum = 0;
                  for(int i = 0 ; i < n ; i++)
                  {
                         arr[i] = sc.nextInt();
                        
                         if(i%2 == 0)
                         {
                                sum += arr[i];
                         arr[i] = -arr[i];
                         }
                  }
                  
                  int arr1[] , arr2[];
                  if(n%2 == 0)
                  {
                  arr1 = new int[n/2];
                  arr2 = new int[n/2 - 1];
                  }
                  
                  else
                  {
                      arr1 = new int[n/2];
                      arr2 = new int[n/2];
                      }
                      
                      int cnt1 = 0 , cnt2 = 0;
                      for(int i = 0 ; i < n ; i++)
                      {
                          if(i%2 == 0)
                          {
                              if(i+1 < n)
                              arr1[cnt1++] = arr[i] + arr[i+1];
                              
                              
                          }
                          
                          else
                          {
                              if(i+1 < n)
                              arr2[cnt2++] = arr[i] + arr[i+1];
                          }
                      }
                      long ml = Math.max(mx(arr1) , mx(arr2)) + sum;
                  System.out.println(Math.max(ml , sum));
                  
              }
       }
}