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 D {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int i=0; i<t; i++) {
           solve(sc);
        }
    }

    static void solve(Scanner sc) {

        int n = sc.nextInt();
        int[] a = new int[n];
        long sum = 0;
        for(int i=0 ;i<n; i++) {
            a[i] = sc.nextInt();
            if(i%2 == 0) sum+= a[i];
        }
        long res = sum;
        long min = 0;
        long max = 0;
        long cur = 0;
        for(int i=0 ;i+1<n; i+=2) {
            cur += a[i+1]-a[i];
            max = Math.max(max, cur-min);
            min = Math.min(min, cur);
        }
        res = Math.max(res, sum + max);

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

        res = Math.max(res, sum + max);

        System.out.println(res);


    }

}