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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.stream.LongStream;
import java.util.HashMap;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.OptionalLong;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DMaximumSumOnEvenPositions solver = new DMaximumSumOnEvenPositions();
        solver.solve(1, in, out);
        out.close();
    }

    static class DMaximumSumOnEvenPositions {
        long cur_sum;
        int start;
        int end;

        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int t = sc.nextInt();
            while (t-- > 0) {
                int n = sc.nextInt();
                long[] arr = new long[n];
                long[] ce = new long[n];
                long[] co = new long[n];
                long[] cce = new long[n];
                long[] cco = new long[n];
                long[] ccdif = new long[n];
                long[] cdif = new long[n];


                for (int i = 0; i < n; i++) {
                    arr[i] = sc.nextInt();
                    if (i % 2 == 0) ce[i] = arr[i];
                    else co[i] = arr[i];
                    if (i == 0) {
                        cco[i] = co[i];
                        cce[i] = ce[i];
                    } else {
                        cco[i] = cco[i - 1] + co[i];
                        cce[i] = cce[i - 1] + ce[i];
                    }
                    cdif[i] = co[i] - ce[i];
                    if (i == 0) {
                        ccdif[i] = cdif[i];
                    } else {
                        ccdif[i] = ccdif[i - 1] + cdif[i];
                    }
                }
                //pw.println(Arrays.toString(ccdif));
                cur_sum = 0;
                start = 0;
                end = -1;
                // pw.println(Arrays.toString(cdif));
                long max = maxEvenLenSum(cdif, n);
                if (max > 0) {
                    subArraySum(cdif, n, max);
                    long ans = 0;
                    if (start > 0) ans += cce[start - 1];
                    if (end != -1) {
                        ans += cce[n - 1] - cce[end];
                        ans += cco[end] - (start > 0 ? cco[start - 1] : 0);
                    }
                    pw.println(ans);
                } else {
                    pw.println(cce[n - 1]);
                }


            }
        }

        public void subArraySum(long[] arr, int n, long sum) {
            //cur_sum to keep track of cummulative sum till that point

            HashMap<Long, Integer> hashMap = new HashMap<>();

            for (int i = 0; i < n; i++) {
                cur_sum = cur_sum + arr[i];
                //check whether cur_sum - sum = 0, if 0 it means
                //the sub array is starting from index 0- so stop
                if (cur_sum - sum == 0) {
                    start = 0;
                    end = i;
                    break;
                }
                //if hashMap already has the value, means we already
                // have subarray with the sum - so stop
                if (hashMap.containsKey(cur_sum - sum)) {
                    start = hashMap.get(cur_sum - sum) + 1;
                    end = i;
                    break;
                }
                //if value is not present then add to hashmap
                hashMap.put(cur_sum, i);

            }
            // if end is -1 : means we have reached end without the sum
//        if (end == -1) {
//            System.out.println("No subarray with given sum exists");
//        } else {
//            System.out.println("Sum found between indexes "
//                    + start + " to " + end);
//        }

        }

        static long maxEvenLenSum(long arr[], int n) {

            // There has to be at
            // least 2 elements
            if (n < 2)
                return 0;

            // dp[i] will store the maximum
            // subarray sum of even length
            // starting at arr[i]
            long[] dp = new long[n];

            // Valid subarray cannot start from
            // the last element as its
            // length has to be even
            dp[n - 1] = 0;
            dp[n - 2] = arr[n - 2] + arr[n - 1];

            for (int i = n - 3; i >= 0; i--) {

                // arr[i] and arr[i + 1] can be added
                // to get an even length subarray
                // starting at arr[i]
                dp[i] = arr[i] + arr[i + 1];

                // If the sum of the valid subarray starting
                // from arr[i + 2] is greater than 0 then it
                // can be added with arr[i] and arr[i + 1]
                // to maximize the sum of the subarray
                // starting from arr[i]
                if (dp[i + 2] > 0)
                    dp[i] += dp[i + 2];
            }

            // Get the sum of the even length
            // subarray with maximum sum
            long maxSum = Arrays.stream(dp).max().getAsLong();
            return maxSum;
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

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

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

    }
}

