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.io.*;
import java.util.*;
import java.util.*;
import java.math.*;
public class Euler {

	public static void main(String[] args) {

		FastReader in = new FastReader();
		PrintWriter o = new PrintWriter(System.out);

		int t = in.nextInt();
		while (t-- > 0) {
			int n = in.nextInt();
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = in.nextInt();
			}
			long[][] dp = new long[n+1][3];
			for (int i = 0; i < n; i++) {
				dp[i+1][0] = Math.max(dp[i+1][0], dp[i][0] + (i % 2 == 0 ? arr[i] : 0));
				if (i + 2 <= n) {
					dp[i+2][1] = Math.max(dp[i+2][1], Math.max(dp[i][0], dp[i][1])) + (i % 2 == 0 ? arr[i+1] : arr[i]);
				}
				dp[i+1][2] = Math.max(dp[i+1][2], Math.max(dp[i][0], Math.max(dp[i][1], dp[i][2]))) + (i % 2 == 0 ? arr[i] : 0);
			}
			o.println(Math.max(dp[n][0], Math.max(dp[n][1], dp[n][2])));
		}

		o.close();
		return;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		int[] readArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}

		long[] readLongArray(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++) a[i] = nextLong();
			return a;
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}