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.*;

public class Program {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        try {
            int t = scanner.nextInt();
            while (t-- > 0) {
                int n = scanner.nextInt(), prev = 0;
                long sum = 0, max = 0, inc_odd = 0, inc_even = 0;
                for (int i=0; i<n; i++) {
                    int p = scanner.nextInt();
                    if ((i & 1) == 0) {
                        sum += p;
                        if (i > 0) {
                            if (inc_even < 0) inc_even += p - prev;
                            else inc_even = p - prev;
                            max = Math.max(max, -inc_even);
                        }
                    } else {
                        if (inc_odd > 0) inc_odd += p - prev;
                        else inc_odd = p - prev;
                        max = Math.max(max, inc_odd);
                    }
                    prev = p;
                }
                System.out.println(sum + max);
            }
        } finally {
            scanner.close();
        }
    }   
    
}