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.BigInteger;
import java.util.*;
public class kadenversion {
    static Scanner sc=new Scanner(System.in);
    public static void main(String[] args) {
        for(int t=sc.nextInt();t>0;t--) {
            show();
        }
    }
    private static void show() {
        int n=sc.nextInt();
        long arr[]=new long [n+1];
        long ans=0;
        long a=0,b=0,max=0;
        for(int i=1;i<=n;i++) {
            arr[i]=sc.nextLong();
            if(i%2==1)ans+=arr[i];
        }
        for(int i=2;i<=n;i+=2) {
            a=Math.max(0, arr[i]-arr[i-1]+a);
            max=Math.max(max,a);
        }
        for(int i=3;i<=n;i+=2) {
            b=Math.max(0, arr[i-1]-arr[i]+b);
            max=Math.max(max, b);
        }
        System.out.println(ans+max);
    }
}