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 D90 {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = sc.nextInt();
        while (t > 0) {
            int n = sc.nextInt();
            int [] a = new int[n]; long sum = 0;
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                if (i % 2 == 0) sum += a[i];
            }
            long [] pref = new long[n+1];
            for (int i = 1; i <= n; i++) {
                if (i % 2 == 1) {
                    pref[i] = pref[i - 1] - a[i - 1];
                } else {
                    pref[i] = pref[i - 1] + a[i - 1];
                }
            }
            long big = Long.MIN_VALUE; long minSoFarOdd = Long.MAX_VALUE / 2; long minSoFarEven = 0;
            for (int i = 1; i <= n; i++) {
                if (i % 2 == 1) {
                    big = Math.max(pref[i] - minSoFarOdd, big);
                    minSoFarOdd = Math.min(minSoFarOdd, pref[i]);
                } else {
                    big = Math.max(big, pref[i] - minSoFarEven);
                    minSoFarEven = Math.min(minSoFarEven, pref[i]);
                }
            }
            out.println(Math.max(sum, sum + big));
            t--;
        }
        out.close();
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }


    }

}