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.*;
import java.io.*;
public class Edu90D {
   public static void main(String[] main) throws Exception{
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer st = new StringTokenizer(br.readLine());
      PrintWriter out = new PrintWriter(System.out);
      int t = Integer.parseInt(st.nextToken());
      for(int trial = 1; trial <= t; trial++){
         st = new StringTokenizer(br.readLine());
         int n = Integer.parseInt(st.nextToken());
         long min1 = 0;
         long min2 = 0;
         long curr1 = 0;
         long curr2 = 0;
         long ans = 0;
         long max = 0;
         int prev = 0;
         st = new StringTokenizer(br.readLine());
         for(int i = 0; i < n; i++) {
        	 int next = Integer.parseInt(st.nextToken());
        	 if(i == 0)
        		 ans += next;
        	 if(i%2 == 0 && i != 0) {
        		 ans += next;
        		 int diff = prev - next;
        		 curr1 += diff;
        		 if(curr1 < min1)
        			 min1 = curr1;
        		 max = Math.max(max,curr1-min1);
        	 }
        	 if(i%2 == 1) {
        		 int diff = next - prev;
        		 curr2 += diff;
        		 if(curr2 < min2)
        			 min2 = curr2;
        		 max = Math.max(max,curr2-min2);
        	 }
        	 prev = next;
         }
         out.println(ans + max);
      }
      out.close();
   }
}