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 _1373_D {
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int t = Integer.parseInt(in.readLine());
		for(int i = 0; i < t; i++) {
			int n = Integer.parseInt(in.readLine());
			StringTokenizer line = new StringTokenizer(in.readLine());
			int[] a = new int[n];
			long sum = 0;
			for(int j = 0; j < n; j++) {
				a[j] = Integer.parseInt(line.nextToken());
				if(j % 2 == 0) sum += a[j]; 
			}
			long[] gain1 = new long[n / 2];
			long[] gain2 = new long[n / 2];
			for(int j = 0; j < n - 1; j += 2) {
				gain1[j / 2] = a[j + 1] - a[j];
			}
			for(int j = 1; j < n - 1; j += 2) {
				gain2[(j - 1) / 2] = a[j] - a[j + 1];
			}
			out.println(sum + Math.max(kadane(gain1), kadane(gain2)));
		}
		in.close();
		out.close();
	}
	static long kadane(long[] gain) {
		long res = 0;
		long sum = 0;
		for(int i = 0; i < gain.length; i++) {
			sum += gain[i];
			if(sum < 0) sum = 0;
			res = Math.max(res, sum);
		}
		return res;
	}
}
