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 MaximumSum {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

		int t = Integer.parseInt(br.readLine());

		while (--t >= 0) {
			int n = Integer.parseInt(br.readLine());
			String[] line = br.readLine().split(" ");
			int[] a = new int[n];
			long sum = 0;
			for (int i = 0; i < n; ++i) {
				a[i] = Integer.parseInt(line[i]);
				if ((i & 1) == 0) {
					sum += a[i];
				}
			}
			int[] difs = new int[n >> 1];
			for (int i = 0, j = 0; j < difs.length; ++i, ++j) {
				difs[j] = -a[i] + a[++i];
			}

			long max_so_far = 0, max_ending_here = 0;

			for (int i = 0; i < difs.length; ++i) {
				max_ending_here += difs[i];
				if (max_so_far < max_ending_here) {
					max_so_far = max_ending_here;
				}

				if (max_ending_here < 0) {
					max_ending_here = 0;
				}
			}

			for (int i = 1, j = 0; j < difs.length; ++j, ++i) {
				difs[j] = a[i];
				if (++i < n) {
					difs[j] -= a[i];
				} else {
					difs[j] = 0;
				}
			}

			max_ending_here = 0;
			for (int i = 0; i < difs.length; ++i) {
				max_ending_here += difs[i];
				if (max_so_far < max_ending_here) {
					max_so_far = max_ending_here;
				}

				if (max_ending_here < 0) {
					max_ending_here = 0;
				}
			}
			sum += max_so_far;

			bw.write(Long.toString(sum));
			bw.newLine();
		}

		br.close();
		bw.close();
	}
}
