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.io.*;
public class Solution {
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int t = Integer.parseInt(st.nextToken());
		PrintWriter out = new PrintWriter(System.out);
		for(int qq=0;qq<t;qq++){
		    st = new StringTokenizer(br.readLine());
		    int n = Integer.parseInt(st.nextToken());
		    st = new StringTokenizer(br.readLine());
		    long[] a = new long[n];
		    for(int i=0;i<n;i++) a[i] = Long.parseLong(st.nextToken());
		    int k = (n+1)/2;
		    long[] bef = new long[k];
		    long[] bes = new long[k];
		    Arrays.fill(bef,0L);
		    Arrays.fill(bes,0L);
		    if(n>=2){
		        bes[0] = Math.max(0L,-a[0] + a[1]);
		    }
		    long res = bes[0];
		    //System.out.println(bef[0]+" "+bes[0]);
		    for(int i=1;i<k;i++){
		        bef[i] = bef[i-1]-a[2*i]+a[2*i-1];
		        bef[i] = Math.max(-a[2*i]+a[2*i-1],bef[i]);
		        bef[i] = Math.max(0,bef[i]);
		        if(2*i+1<n){
		            bes[i] = bes[i-1]-a[2*i]+a[2*i+1];
		            bes[i] = Math.max(-a[2*i]+a[2*i+1],bes[i]);
		            bes[i] = Math.max(0,bes[i]);
		        }
		        res = Math.max(bes[i],res);
		        res = Math.max(bef[i],res);
		        //System.out.println(bef[i]+" "+bes[i]);
		    }
		    for(int i=0;i<n;i+=2) res += a[i];
		    out.println(res);
		}
		out.flush();
	}
}