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 D1371 {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int cases = scanner.nextInt();
		for(int i = 0; i < cases; i++) {
			int length = scanner.nextInt();
			int[] numbers = new int[length];
			long result = 0;
			for(int j = 0; j < length; j++)  {
				numbers[j] = scanner.nextInt();
				if(j % 2 == 0)
					result += numbers[j];
			}
			
			long best = 0;
			long sum = 0;
			
			for (int j = 1; j < length; j += 2) {
				sum = Math.max(0, sum + numbers[j] - numbers[j - 1]);
				best = Math.max(best, sum);
			}
			
			sum = 0;
			for (int j = 1; j < length - 1; j += 2) {
				sum = Math.max(0, sum + numbers[j] - numbers[j + 1]);
				best = Math.max(best, sum);
			}
			
			System.out.println(best + result);
		}
	}
}
