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.lang.*;

public class Main {
    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()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
	
    static FastReader in = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
	public static void solve() {
		int n=in.nextInt();
		long[][] dp = new long[n+1][3];
		int[] arr = new int[n+1];
		for(int i=0;i<n;i++) {
			arr[i] = in.nextInt();
		}
		for(int i=0;i<n;i++) {
			dp[i+1][0] = Math.max(dp[i][0],dp[i+1][0])+ ((i+1)%2)*arr[i];
			if(i+2<=n)
				dp[i+2][1] = Math.max(dp[i][0],Math.max(dp[i+2][1],dp[i][1])) + ((i+1)%2)*arr[i+1] + (i%2)*arr[i];
			dp[i+1][2] = Math.max(Math.max(dp[i][0],dp[i+1][2]),Math.max(dp[i][1],dp[i][2])) + ((i+1)%2)*arr[i];
		}
		out.println(Math.max(dp[n][0],Math.max(dp[n][1],dp[n][2])));
		out.flush();
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int T = 1;
		T = in.nextInt();
		for(int i=1;i<=T;i++) {
//			out.println("Test Case #"+i);
//			out.flush();
			solve();
		}
	}

}
