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 MaximumSumOnEvenPositions {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for (int t = 0; t < T; t++) {
            int N = in.nextInt();
            int[] arr = new int[N];
            long sum = 0;
            for (int i = 0; i < N; i++) {
                arr[i] = in.nextInt();
                if (i % 2 == 0) {
                    sum += arr[i];
                }
            }
            long[] dp = new long[N];
            long max = 0;
            for (int i = 1; i < N; i++) {
                dp[i] = i % 2 == 0 ? arr[i - 1] - arr[i] : arr[i] - arr[i - 1];
                if (i > 2 && dp[i - 2] > 0) {
                    dp[i] += dp[i - 2];
                }
                max = Math.max(max, dp[i]);
            }
            System.out.println(sum + max);
        }
    }
}