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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;



public class D {
	public static long maxSum(int a[]) {
		long curr=a[0],max=a[0];
		for(int i=1;i<a.length;i++) {
			curr=Math.max(a[i], curr+a[i]);
			max=Math.max(curr, max);
		}
		return  max;
		
		
	}

	public static void main(String[] args) {
		 FastScanner sc =new FastScanner();
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			int a[]=sc.readArray(n);
			if(n==1) System.out.println(a[0]);
			else	if(n==2) System.out.println(Math.max(a[0], a[1]));
			else {
				int aeven[]=new int[n/2];
				int k=0;
				// even array
				for(int i=0;i<Math.ceil((double)(n-1)/2);i++) {
					//System.out.println(a[2*i+1]+" "+a[2*i]);
					aeven[k++]=a[2*i+1]-a[2*i];
					
				}
//				for(int i:aeven) System.out.print(i+" ");
//				System.out.println();
				k=0;
				int aodd[]=new int[(n-1)/2];
				for(int i=1;i<=(n-1)/2;i++) {
					aodd[k++]=a[2*i-1]-a[2*i];
				}
				
//				for(int i:aodd) System.out.print(i+" ");
//				System.out.println();
				long sum=0;
				for(int i=0;i<n;i++) sum+= (i%2==0)? a[i]:0;
				
				long max=Math.max(maxSum(aodd), maxSum(aeven));
				long res=Math.max(sum, sum+max);
				System.out.println(res);
				
			}
			
			
			
		}
		    	
		    }
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
}

		    
		}