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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class D {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(System.out);
	private static final String Y = "YES";
	private static final String N = "NO";

	static void solve() throws Exception {
		int tests = nextInt();
		for (int t = 0; t < tests; t++) {
			int n = nextInt();
			long[] nums = new long[n];
			long[] dif0 = new long[n / 2];
			long[] dif1 = new long[(n - 1) / 2];
			long suma = 0;
			for (int i = 0; i < n; i++) {
				nums[i] = nextInt();
				if (i % 2 == 0) {
					suma += nums[i];
					if (i == 0) {
						continue;
					}
					dif1[(i - 1) / 2] = nums[i - 1] - nums[i];
				} else {
					dif0[i / 2] = nums[i] - nums[i - 1];
				}
			}

			out.println(Math.max(getMax(dif0), getMax(dif1)) + suma);
		}
	}

	private static long getMax(long[] difs) {
		long max = 0;
		long cur = 0;
		for (int i = 0; i < difs.length; i++) {
			cur += difs[i];
			if (cur < 0) {
				cur = 0;
			} else {
				max = Math.max(max, cur);
			}
		}
		return max;
	}

	public static void main(String[] args) throws Exception {
		solve();
		out.close();
	}

	static String next() throws Exception {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}

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

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