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 d {
	static final FS sc = new FS();
	static final PrintWriter pw = new PrintWriter(System.out);
	
	static int n;
	static int[] arr;
	static long[] preE, preO;
	public static void main(String[] args) {
		int t = sc.nextInt();
		for(int tt = 1; tt <= t; ++tt) {
			n = sc.nextInt();
			arr = new int[n];
			preE = new long[n];
			preO = new long[n];
			
			for(int i = 0; i < n; ++i) {
				arr[i] = sc.nextInt();
				
				if(i % 2 == 0) preE[i] = arr[i];
				else preO[i] = arr[i];
				
				if(i > 0) {
					preE[i] += preE[i - 1];
					preO[i] += preO[i - 1];
				}
			}
			
			long maxAdd = 0;
			long minEIdxVal = -preE[0], minOIdxVal = 0;
			for(int i = 0; i < n; ++i) {
				long currVal = preO[i] - preE[i];
				long curr = currVal;
				
				if(i % 2 == 0) curr -= minEIdxVal;
				else curr -= minOIdxVal;
				
				maxAdd = Math.max(maxAdd, curr);
				
				if(i % 2 == 0) minEIdxVal = Math.min(minEIdxVal, currVal);
				else minOIdxVal = Math.min(minOIdxVal, currVal);
			}
			
			long out = preE[n - 1] + maxAdd;
			
			pw.println(out);
		}
		pw.flush();
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}