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
//package maximumsumonevenpositions;

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

public class maximumsumonevenpositions {
	
	static long[] b;
	
	public static long maxSubarray() {
		long max = Long.MIN_VALUE / 2;
		long curMax = Long.MIN_VALUE / 2;
		for(int i = 0; i < b.length; i++) {
			curMax = Math.max(curMax + b[i], b[i]);
			max = Math.max(max, curMax);
			//System.out.print(b[i] + " ");
		}
		//System.out.println();
		return max;
	}
	
	public static void main(String[] args) throws IOException {
		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(fin.readLine());
		StringBuilder fout = new StringBuilder();
		while(t-- > 0) {
			int n = Integer.parseInt(fin.readLine());
			long[] nums = new long[n];
			StringTokenizer st = new StringTokenizer(fin.readLine());
			for(int i = 0; i < n; i++) {
				nums[i] = Integer.parseInt(st.nextToken());
			}
			b = new long[n / 2];
			long ans = 0;
			for(int i = 0; i < n; i++) {
				if(i % 2 == 0) {
					ans += nums[i];
					if(i + 1 != n) {
						b[i / 2] = nums[i + 1] - nums[i];
					}
				}
			}
			//System.out.println("CUR ANS: " + ans);
			long add = maxSubarray();
			if(n % 2 == 1) {
				b = new long[n / 2];
				for(int i = n - 1; i > 0; i--) {
					if(i % 2 == 0) {
						b[i / 2 - 1] = nums[i - 1] - nums[i];
					}
				}
				add = Math.max(add, maxSubarray());
			}
			else {
				b = new long[n / 2 - 1];
				for(int i = n - 2; i > 0; i--) {
					if(i % 2 == 0) {
						b[i / 2 - 1] = nums[i - 1] - nums[i];
					}
				}
				add = Math.max(add, maxSubarray());
			}
			//System.out.println("ADD: " + add);
			ans = Math.max(ans, ans + add);
			fout.append(ans).append("\n");
		}
		System.out.print(fout);
	}
}
