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.nextInt();
// gatogari.
public class Solution {
    public static void main(String[] args) {       

        Scanner scanner = new Scanner(System.in);
        int test_case = scanner.nextInt();
            
            for(int z = 0; z < test_case; ++z)
            {
                int n = scanner.nextInt();
		int[] array = new int[n];
                

                for(int i = 0; i < n; ++i)
                    {
                        array[i] = scanner.nextInt();
                    }
                 
                long[][] dp = new long[n+1][3];
                
                for(int i = 0; i < n; ++i)
                {
                    dp[i+1][0] = Math.max(dp[i+1][0], dp[i][0] + (i % 2 == 0 ? array[i] : 0));
                    if(i+2 <= n) dp[i+2][1] = Math.max(dp[i+2][1], Math.max(dp[i][0], dp[i][1]) + (i % 2 == 0 ? array[i+1] : array[i]));
                    dp[i+1][2] = Math.max(dp[i+1][2], Math.max(dp[i][0], Math.max(dp[i][1], dp[i][2])) + (i % 2 == 0 ? array[i] : 0));
                    
                }
                System.out.println(Math.max(dp[n][0], Math.max(dp[n][2], dp[n][1])));
                //String[] arr = {"a","b","c", "d","e", "f" , "g" , "h", "i" , "j" , "k" , "l", "m", "n", "o", "p", "q" , "r", "s", "t", "u" , "v", "w" ,"x", "y", "z"};
 
            }
                    
    }
}

