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.io.PrintWriter;
import java.util.StringTokenizer;

public class NifflerD {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        PrintWriter pw = new PrintWriter(System.out);
        
        try {
            int t = Integer.parseInt(br.readLine());
            while(t-->0) {
                int n = Integer.parseInt(br.readLine());
                int[] a = new int[n];
                st = new StringTokenizer(br.readLine());
                long initsum = 0;
                for(int i=0 ; i<n ; i++) {
                    a[i] = Integer.parseInt(st.nextToken());
                    if(i%2 == 0) {
                        initsum += a[i];
                    }
                }

                long sum = 0;
                long max = 0;
                for(int i=0 ; i<n ; i+=2) {
                    if(i+1 < n) {
                        sum += a[i+1]-a[i];
                        max = Math.max(max, sum);
                        if(sum < 0) {
                            sum = 0;
                        }
                    }
                }

                long first = initsum + max;

                max = 0;
                sum = 0;
                for(int i=1 ; i<n ; i+=2) {
                    if(i+1 < n) {
                        sum += a[i] - a[i+1];
                        max = Math.max(max, sum);
                        if(sum < 0) {
                            sum = 0;
                        }
                    }
                }

                long sec = initsum + max;

                long ans = Math.max(first, sec);

                pw.println(ans);
            }
        }
        finally {
            pw.flush();
            pw.close();
        }
    }
}
