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.io.*;
import java.util.*;

public class swap {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
		
		int t = Integer.parseInt(br.readLine());
		
		for (int i = 0; i < t; i++){
			int n = Integer.parseInt(br.readLine());
			StringTokenizer st = new StringTokenizer(br.readLine());
			long sum = 0;
			long[] arr = new long[n];
			
			for (int j = 0; j < n; j++){
				arr[j] = Integer.parseInt(st.nextToken());
				if (j % 2 == 0) sum += arr[j];
			}
			
			long[] dp = new long[n + 3];
			if (n >= 2) dp[1] = Math.max(0, arr[1] - arr[0]);
			if (n >= 3) dp[2] = Math.max(0, arr[1] - arr[2]);
			long max = Math.max(dp[1], dp[2]);
			
			for (int j = 3; j < n; j++){
				if (j % 2 == 1){
					dp[j] = Math.max(0, arr[j] - arr[j - 1] + dp[j - 2]);
					max = Math.max(max, dp[j]);
				}
				else {
					dp[j] = Math.max(0, arr[j - 1] - arr[j] + dp[j - 2]);
					max = Math.max(max, dp[j]);
				}
			}
			
			pw.println(sum + max);
		}
		
		br.close();
		pw.close();
	}
}
