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.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;

public class Solution {
	public static class FastReader {
			BufferedReader br;
			StringTokenizer root;
			
	 
			public FastReader() {
				br = new BufferedReader(new InputStreamReader(System.in));
			}
	 
			String next() {
				while (root == null || !root.hasMoreTokens()) {
					try {
						root = new StringTokenizer(br.readLine());
					} catch (Exception addd) {
						addd.printStackTrace();
					}
				}
				return root.nextToken();
			}
	 
			int nextInt() {
				return Integer.parseInt(next());
			}
	 
			double nextDouble() {
				return Double.parseDouble(next());
			}
	 
			long nextLong() {
				return Long.parseLong(next());
			}
	 
			String nextLine() {
				String str = "";
				try {
					str = br.readLine();
				} catch (Exception addd) {
					addd.printStackTrace();
				}
				return str;
			}
		}
	 
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	public static FastReader sc = new FastReader();
 
	static long mod = (int) (1000000000),MAX=(int)(1e5+10);
	
	static int[][] dp;
	
	public static void main(String[] args) throws IOException{
	    
		int t = sc.nextInt();
		while(t-->0) {
			int n = sc.nextInt();
			int[] a = new int[n];
			long ans = 0;
			for(int i=0;i<n;++i) {
				a[i] = sc.nextInt();
				if((i&1) == 0) {
					ans+=a[i];
					a[i] = -a[i];
				}
			}
			long sum = 0,max = 0,minEven = 0,minOd = 0;
			for(int i=0;i<n;++i) {
				sum+=a[i];
				if((i&1) == 0) {
					minEven = Math.min(sum, minEven);
					max = Math.max(max, sum-minEven);
				}else {
					minOd = Math.min(sum, minOd);
					max = Math.max(max, sum-minOd);
				}
			}
			ans+=max;
			out.println(ans);
	     }
		 out.close();
	
	
}
}



