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.*;
import java.io.*;

public class D{
    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;
        }
    }

    static int gcd(int a, int b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static int lcm(int a, int b) {
        return (a * b) / gcd(a, b);
    }

    static long maxSubArraySum(long a[], int size) {
        long max_so_far = Long.MIN_VALUE, max_ending_here = 0, start = 0, end = 0, s = 0;

        for (int i = 0; i < size; i++) {
            max_ending_here += a[i];

            if (max_so_far < max_ending_here) {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }

            if (max_ending_here < 0) {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        return max_so_far;
    }

    public static void main(String[] args) {
        FastReader s = new FastReader();
        int t = s.nextInt();
        while (t-- > 0) {
            int n = s.nextInt();

            long a[] = new long[n];
            long mysum = 0;

            for(int i= 0;i<n;i++){
                a[i] = s.nextLong();
                if(i%2==0)
                mysum+=a[i];
            }

            if(n%2==0){
                long sum1[] = new long[n/2];
                
                for(int i=0;i<n;i+=2){
                    sum1[i/2] = a[i+1] - a[i];
                }

                long max1 = maxSubArraySum(sum1, n/2);

                long sum2[] = new long[(n/2)-1];

                for(int i=1;i<n-1;i+=2){
                    sum2[(i-1)/2] = a[i] - a[i+1];
                }

                long max2 = maxSubArraySum(sum2, (n/2)-1);
                long max =Math.max(max1,max2);
                // if(n==200000)
                // System.out.println(mysum + " "+ max);
                if(max>0)
                System.out.println(max+mysum);
                else
                System.out.println(mysum);
            }

            else{
                long first[] = new long[(n-1)/2];
                long second[] = new long[(n - 1) / 2];

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

                long max1 = maxSubArraySum(first, (n-1)/2);

                for(int i=1;i<n;i+=2){
                    second[(i-1)/2] = a[i] - a[i+1];
                }
                long max2 = maxSubArraySum(second, (n-1)/2);

                long max = Math.max(max1,max2);

                if(max>0)
                System.out.println(mysum+max);
                else
                System.out.println(mysum);
            }

        }
    }
}