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.util.Map.Entry;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;


public class CF {
	private static FS sc = new FS();
	
	private static class FS {
		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) {}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	private static class extra {
		
		static int[] intArr(int size) {
			int[] a = new int[size];
			for(int i = 0; i < size; i++) a[i] = sc.nextInt();
			return a;
		}
		
		static long[] longArr(int size) {
			long[] a = new long[size];
			for(int i = 0; i < size; i++) a[i] = sc.nextLong();
			return a;
		}
		
		static long intSum(int[] a) {
			long sum = 0; 
			for(int i = 0; i < a.length; i++) {
				sum += a[i];
			}
			return sum;
		}
		
		static long longSum(long[] a) {
			long sum = 0; 
			for(int i = 0; i < a.length; i++) {
				sum += a[i];
			}
			return sum;
		}
		
		static LinkedList[] graphD(int vertices, int edges) {
			LinkedList<Integer>[] temp = new LinkedList[vertices+1];
			for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
			for(int i = 0; i < edges; i++) {
				int x = sc.nextInt();
				int y = sc.nextInt();
				temp[x].add(y);
			}
			return temp;
		}
		
		static LinkedList[] graphUD(int vertices, int edges) {
			LinkedList<Integer>[] temp = new LinkedList[vertices+1];
			for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
			for(int i = 0; i < edges; i++) {
				int x = sc.nextInt();
				int y = sc.nextInt();
				temp[x].add(y);
				temp[y].add(x);
			}
			return temp;
		}
		
		static void printG(LinkedList[] temp) {
			for(LinkedList<Integer> aa:temp) System.out.println(aa);
		}
		
		static long cal(long val, long pow) {
			if(pow == 0) return 1;
			long res = cal(val, pow/2);
			
//			long ret = (res*res)%mod;
//			if(pow%2 == 0) return ret;
//			return (val*ret)%mod;
			long ret = res*res;
			if(pow%2 == 0) return ret;
			return val*ret;
		}
		
		static long gcd(long a, long b) { return b == 0 ? a:gcd(b, a%b); }
		
	}
	
	static int mod = (int) 1e9 + 7;
//	static int mod = (int) 998244353;
	static int max = (int) 1e7+1;
//	static LinkedList<Integer>[] temp;
//	static int[] dep, dia;
	
	public static void main(String[] args) {	
		int t = sc.nextInt();
//		int t = 1;
		StringBuilder ret = new StringBuilder();
		while(t-- > 0) {
			int n = sc.nextInt();
			int[] a = extra.intArr(n);
			long max = 0, cur = 0;
			for(int i = 0; i < n-1; i += 2) {
				cur += a[i+1]-a[i];
				cur = Math.max(0, cur);
				max = Math.max(cur, max);
			}
			cur = 0;
			for(int i = 1; i < n-1; i += 2) {
				cur += a[i]-a[i+1];
				cur = Math.max(0, cur);
				max = Math.max(cur, max);
			}
			long sum = 0;
			for(int i = 0; i < n; i += 2) sum += a[i];
			ret.append((sum+max) + "\n");
		}
		System.out.println(ret);
	}
}