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.Arrays;
import java.util.Scanner;

public class Solution {

    private static long findMax(long[] a) {
        int len = a.length;

        long[] prefix = new long[len + 1];

        for (int i = 0; i < len; i += 1) {
            prefix[i + 1] = prefix[i] + a[i];
        }

        long[] min = new long[len + 1];

        for (int i = 0; i < len; i += 1) {
            min[i + 1] = Math.min(min[i], prefix[i + 1]);
        }

        long max = 0;

        for (int i = 0; i < len; i += 1) {
            max = Math.max(max, prefix[i + 1] - min[i + 1]);
        }

        return max;
    }

    public static void main(String[] args) {
        final Scanner in = new Scanner(System.in);

        final int T = in.nextInt();

        for (int t = 0; t < T; t += 1) {
            final int n = in.nextInt();

            final long[] a = new long[n];

            for (int i = 0; i < n; i += 1) a[i] = in.nextLong();

            long sum = 0;
            for (int i = 0; i < n; i += 2) sum += a[i];

            final long[] even = new long[n / 2];

            for (int i = 0; i + 1 < n; i += 2) {
                even[i / 2] = a[i + 1] - a[i];
            }

            final long[] odd = new long[n / 2];

            for (int i = 1; i + 1 < n; i += 2) {
                odd[(i - 1) / 2] = a[i] - a[i + 1];
            }

//            System.out.println(Arrays.toString(a));
//            System.out.println(Arrays.toString(even));
//            System.out.println(Arrays.toString(odd));

            System.out.println(sum + Math.max(findMax(even), findMax(odd)));
        }
    }
}
