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.*;
import java.text.*;

public class Main {
    static long[][][]memo;
    static int n,a[];
    static long dp(int done,int idx,int par){
        if (idx==n)
            return done==1?(long)-1e16:0;
        if (memo[par][done][idx]!=-1)
            return memo[par][done][idx];
        long ans;
        ans=(par==0?a[idx]:0)+dp(done,idx+1,1^par);
        if (done==0){
            if (par==0){
                ans=Math.max(ans,dp(1,idx+1,par));
            }
        }
        else if (done==1){
            if (par==0){
                ans=Math.max(ans,dp(2,idx+1,par)+a[idx]);
            }
        }
      return memo[par][done][idx]=ans;
    }
    static long dp2(int done,int idx,int par){
        if (idx==n)
            return done==1?(long)-1e16:0;
        if (memo[par][done][idx]!=-1)
            return memo[par][done][idx];

        long ans=(par==0?a[idx]:0)+dp2(done,idx+1,par^1);
        if (done==0&&par==1&&idx+1<n){
            ans=Math.max(ans,dp2(1,idx+1,1)+a[idx]);
        }
        if (done==1&&par==1){
            ans=Math.max(ans,dp2(2,idx+1,1));
        }
        return memo[par][done][idx]=ans;
    }
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        while (t-->0){
            n = sc.nextInt();
            a= new int[n];
            for (int i =0;i<n;i++)a[i]=sc.nextInt();
            memo= new long[2][3][n];
            for (long[][]x:memo)for (long[]z:x)Arrays.fill(z,-1);
            long ans=dp(0,0,0);
            for (long[][]x:memo)for (long[]z:x)Arrays.fill(z,-1);
            ans=Math.max(ans,dp2(0,0,0));
            pw.println(ans);
        }
        pw.flush();
    }
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

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

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

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

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

        public int[] nextIntArr(int n) throws IOException {
            int[] arr = new int[n];
            for (int i = 0; i < arr.length; i++)
                arr[i] = nextInt();
            return arr;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }
}