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 idk{
public static void main(String[] args)
{

   Scanner myObj = new Scanner(System.in);
   int cases = myObj.nextInt();
   int size;
   int[] stuffs = new int[0];
   int[] a1 = new int[0];
   int[] a2 = new int[0];
   long[] answers = new long[cases];
   long sum = 0;
   for(int i = 0; i < cases; i++)
   {
      size = myObj.nextInt();
      stuffs = new int[size];
      for(int j = 0; j < size; j++)
      {
         stuffs[j] = myObj.nextInt();
      }
      a1 = new int[size/2];
      a2 = new int[(size-1)/2];
      for(int j = 0; j < size/2; j++)
      {
         a1[j] = stuffs[2*j+1] - stuffs[2*j];
      }
      for(int j = 0; j < (size-1)/2; j++)
      {
         a2[j] = stuffs[2*j+1] - stuffs[2*j+2];
      }
      sum = 0;
      for(int j = 0; j < (size+1)/2; j++)
      {
         sum += stuffs[2*j];
      }
      answers[i] = sum + Math.max(kadane(a1), kadane(a2));
   }
   for(int i = 0; i < cases; i++)
   {
      System.out.println(answers[i]);
   }
}
public static long kadane(int[] arr)
{
   long[] sums = new long[arr.length+1];
   sums[0] = 0;
   for(int i = 1; i < arr.length+1; i++)
   {
      sums[i] = Math.max(arr[i-1], arr[i-1]+sums[i-1]);
   }
   long max = 0;
   for(int i = 0; i < arr.length+1; i++)
   {
      max = Math.max(max, sums[i]);
   }
   return max;
}

}