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.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;

public class codechampscodechef {

 static   class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void main(String[] args) {
        FastReader s = new FastReader();
        int a = s.nextInt();
        StringBuilder sb=new StringBuilder();
        for (int i = 0; i < a; i++) {
            int b = s.nextInt();
            int arr[] = new int[b];
            long currsum = 0;
            for (int j = 0; j < b; j++) {
                arr[j] = s.nextInt();
                if (j % 2 == 0)
                    currsum += arr[j];
            }
            long maxdiff = 0;
            int j=2;
            if(b==1);
            else {
                long dp[] = new long[b];
                long odd = 0, even = 0;
                for (int k = 0; k < j; k++) {
                    if (k % 2 == 0)
                        even += arr[k];
                    else
                        odd += arr[k];
                }
                dp[0] = odd - even;
                maxdiff = Math.max(maxdiff, odd - even);
                for (int k = 1; k <= b - j; k++) {
                    if ((k - 1) % 2 == 0) {
                        even -= arr[k - 1];
                        even += arr[k + j - 1];
                    } else {
                        odd -= arr[k - 1];
                        odd += arr[k + j - 1];
                    }
                    dp[k] = odd - even;
                }
                long sum = 0;
                for (int k = 0; k < b; k += 2) {
//                    System.out.print(dp[k]+" ");
                    sum = Math.max(dp[k], sum + dp[k]);
//                    System.out.println(sum+" "+k);
                    maxdiff = Math.max(sum, maxdiff);
                }
                long sum1 = 0;
//            System.out.println(maxdiff);
                for (int k = 1; k < b; k += 2) {
                    sum1 = Math.max(dp[k], sum1 + dp[k]);
                    maxdiff = Math.max(sum1, maxdiff);
                }
            }
//            System.out.println(maxdiff);
            sb.append((currsum+maxdiff)+"\n");
        }
        System.out.println(sb.toString());
    }
}
