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.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner (System.in);
        int t = sc.nextInt();
        int a [] ;
        long f [][];
        long sum;
        // process each test
        for (int x = 0 ; x < t ; x++){
            int n = sc.nextInt();
            // new init
            a     = new int [n+10];
            f     = new long [n+10][2] ;
            sum   = 0 ;

            for (int  i = 0 ; i < n ; i++ )
            {
                a[i] = sc.nextInt();
                if (i % 2 ==0) sum += a[i] ;
            }

            for (int  i = 0; i <n ; i+=2){
                f[i][0] = 0 ;
                if (i-2 >= 0 ) {
                    f[i][0] = Math.max(f[i-2][0]- a[i]+ a[i-1],f[i][0]);
                    if (i+1 < n ) f[i][1] = Math.max (f[i-2][1] - a[i] + a[i+1], f[i][1]);
                }
                if (i > 0 ) f[i][0] = Math.max(f[i][0], -a[i] + a[i-1]);
                if (i + 1  < n ) f[i][1] = Math.max (f[i][1], -a[i]+a[i+1]);

            }
            long maxi = Integer.MIN_VALUE;
            for (int  i  = 0 ; i < n ; i+=2)  maxi = Math.max (maxi,Math.max(f[i][0],f[i][1]));
            System.out.println (sum + maxi);
        }

    }
}
