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 scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while ((t--) > 0){
            int n = scanner.nextInt();
            int[] a = new int[n];
            for(int i=0; i<n; ++i)a[i] = scanner.nextInt();
            System.out.println(maxSum(a));

        }
    }

    public static long maxSum(int[] a){
        // 求偶数和
        long sum = 0;
        for(int i=0; i<a.length; ++i){
            sum += ((i & 1) == 0 ? a[i] : 0);
        }

        // 求偶数位与前一位交换 sum(连续子数组交换后奇数减偶数的最大值)
        long acc = 0;
        long max = 0;
        for(int i=2; i<a.length; i+=2){
            if(acc < 0){
                acc = a[i-1] - a[i];
            }else{
                acc += a[i-1] - a[i];
            }

            max = Math.max(acc, max);
        }

        // 求奇数与其前一位交换
        acc = 0;
        for(int i=1; i<a.length; i+=2){
            if(acc < 0){
                acc = a[i] - a[i-1];
            }else{
                acc += a[i] - a[i-1];
            }
            max = Math.max(acc, max);
        }

        // 初始偶数位和 + 连续区间里（奇数和 - 偶数和）（交换后奇数位变为偶数位）
        return sum + max;
    }
}
