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.*;
import java.util.*;
public class Q4 {
 
	public static void main(String[] args) { 
		FastScanner input = new FastScanner();
		PrintWriter w = new PrintWriter(System.out);
		int T = input.nextInt();
		while (T-->0) {
			int N = input.nextInt();
			long[] a = new long[N];
			long ans = 0;
			for (int j = 0; j < N; j++) {
				a[j] = input.nextLong();
				if (j%2==0) ans+=a[j];
			} 
			//Optimize
			/*w.println(ans);
			w.flush();*/
			long cur = 0;
			long bestup = 0;
			for (int j = 0; j < N-1; j+=2) {
				cur+=a[j+1]-a[j];
				cur=Math.max(cur, 0);
				bestup=Math.max(bestup, cur);
			}
			cur = 0; //Refresh potential answer
			for (int j = 1; j < N-1; j+=2) { //7 8 4 --> 7 4 8 is a better possibility
				cur+=a[j]-a[j+1];
				cur=Math.max(cur, 0);
				bestup=Math.max(cur, bestup);
			}
			ans+=bestup;
			w.println(ans);
			w.flush();
		}
	}
	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());
		}
	}
}
