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.Scanner;

public class Problem4 {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		in.nextLine();
		for (int i = 0; i < t; i++) {
			solve(in);
		}
	}

	private static void solve(Scanner in) {
		int n = in.nextInt();
		
		int arr[] = new int[n];
		long sum = 0; 
		for (int i = 0; i < n; i++) {
			arr[i] = in.nextInt();
			
			if (i % 2 == 0) {
				sum += arr[i];
			}
		}
		
		long[] dp = new long[n + 1];
		
		for (int i = 2; i <= n; i++) {
			if (i % 2 == 1) {
				long diff = arr[i - 2] - arr[i - 1];
				dp[i] = Math.max(0, dp[i-2] + diff);
			} else {
				long diff = arr[i - 1] - arr[i - 2];
				dp[i] = Math.max(0, dp[i-2] + diff);
			}
		}
		
		long max = 0;
		for (int i = 0; i <= n; i++) {
			max = Math.max(dp[i], max);
		}
		System.out.println(sum + max);
	}	
}
