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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

/**
 * @project: CodeforcesEduc90
 * @author: izhanatuly
 * @date: 25/06/2020
 */
public class D {
    static Scanner scanner;

    public static void main(String[] args) {
        scanner = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int cases = scanner.nextInt();
        for (int t = 0; t < cases; t++) {
            runTestCase();
        }
    }

    public static void runTestCase() {
        int n = scanner.nextInt();
        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = scanner.nextInt();
        }
        long var1 = cadane(nums, true);
        long var2 = cadane(nums, false);
        System.out.println(Math.max(var1, var2));
    }

    static long cadane(int[] nums, boolean cmpBack) {
        int n = nums.length;
        // calc sum on evens
        long sumEvens = 0;
        for(int i=0; i<n; i++) {
            if(i%2==0) sumEvens += nums[i];
        }
        // calc diff on odds with cadane
        long currSum = 0;
        long bestSum = 0;
        for(int i=1; i<n; i+=2) {
            long diff;
            if(cmpBack) {
                diff = nums[i] - nums[i-1];
            } else {
                if(i<n-1) {
                    diff = nums[i] - nums[i+1];
                } else {
                    diff = 0;
                }
            }
            if(currSum+diff < diff) {
                currSum = diff;
            } else {
                currSum += diff;
            }
            if(currSum > bestSum) {
                bestSum = currSum;
            }
        }
        // return best cadane + sum on evens
        return bestSum + sumEvens;
    }
}
