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.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class d {

	public static void main(String[] args) throws Exception {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		sc = new FastScanner(in);
		int t = sc.nextInt();
	   

		while (t-- > 0) {
			int n = sc.nextInt();
			long[] arr = new long[n];
			long sum = 0 ;
			for(int i = 0 ;i < n;i++) {
				arr[i] = sc.nextLong();
				if(i%2==0) {
					sum+=arr[i];
				}
			}
			long sero = kadane(arr , 0  , n);
			long one = kadane2(arr , 1 , n);
			out.println(sum+Math.max(sero, one));
		}

		out.flush();

	}
    
	private static long kadane(long[] arr, int i, int n) {
		long[] temp = new long[n/2];
		int ind =0 ;
		while(i < n-1) {
			long even   = arr[i];
			long odd = arr[i+1];
			i+=2;
			long diff = even - odd;
			temp[ind]=diff;
		ind++;
		}
		long min = Long.MAX_VALUE;
		long cur = 0;
		for( i = 0 ; i < ind ;i++) {
			cur+=(long)temp[i];
			if(cur>0) {
				cur=0;
			}
			if(cur < min) {
				min = cur;
			}
		}
		
		return (min == Long.MAX_VALUE)?0 : min*-1;
		
	}
	private static long kadane2(long[] arr, int i, int n) {
		long[] temp = new long[n/2];
		int ind =0 ;
		while(i < n-1) {
			long even   = arr[i+1];
			long odd = arr[i];
			i+=2;
			long diff = even - odd;
			temp[ind]=diff;
		ind++;
		}
		long min = Long.MAX_VALUE;
		long cur = 0;
		for( i = 0 ; i < ind ;i++) {
			cur+=(long)temp[i];
			if(cur>0) {
				cur=0;
			}
			if(cur < min) {
				min = cur;
			}
		}
		
		return (min == Long.MAX_VALUE)?0 : min*-1;
		
	}

	static BufferedReader in;
	static FastScanner sc;
	static PrintWriter out;
	static long mod = 1000000009L;
	static class FastScanner {
		BufferedReader in;
		StringTokenizer st;

		public FastScanner(BufferedReader in) {
			this.in = in;
		}

		public String nextToken() throws Exception {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(in.readLine());
			}
			return st.nextToken();
		}

		public int nextInt() throws Exception {
			return Integer.parseInt(nextToken());
		}

		public long nextLong() throws Exception {
			return Long.parseLong(nextToken());
		}

		public double nextDouble() throws Exception {
			return Double.parseDouble(nextToken());
		}
	}
}
