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
//created by Whiplash99
import java.io.*;
import java.util.*;
public class D
{
    private static long INF=(long)(-1e16);
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N;

        int T=Integer.parseInt(br.readLine().trim());
        StringBuilder sb=new StringBuilder();

        while(T-->0)
        {
            N=Integer.parseInt(br.readLine().trim());
            String s[]=br.readLine().trim().split(" ");
            long a[]=new long[N];
            for(i=0;i<N;i++) a[i]=Integer.parseInt(s[i]);

            long b[]=new long[N];
            for(i=0;i<N;i++)
            {
                b[i]=a[i];
                if(i%2==0) b[i]=-a[i];
            }

            long sum=0;
            for(i=0;i<N;i+=2) sum+=a[i];

            long max=0,cur=0,tmp=0;
            int l=-1,r=-1;

            for(i=0;i<N-1;i+=2)
            {
                tmp=b[i]+b[i+1];
                cur=Math.max(cur+tmp,tmp);
                if(cur>max)
                {
                    max=cur;
                    r=i+1;
                }
            }

            cur=0;
            for(i=r;i>=0;i--)
            {
                cur+=b[i];
                if(cur==max&&(r-i+1)%2==0) break;
            }
            l=i;

            long max2=0,cur2=0;
            int l2=-1,r2=-1;

//            for(i=0;i<N;i++)
//            {
//                b[i]=a[i];
//                if(i%2==1) b[i]=-a[i];
//            }

            for(i=1;i<N-1;i+=2)
            {
                tmp=b[i]+b[i+1];
                cur2=Math.max(cur2+tmp,tmp);
                if(cur2>max2)
                {
                    max2=cur2;
                    r2=i+1;
                }
            }

            cur2=0;
            for(i=r2;i>=0;i--)
            {
                cur2+=b[i];
                if(cur2==max2&&(r2-i+1)%2==0) break;
            }
            l2=i;

            long sum2=0,sum3=0;

            if(r!=-1&&l!=-1)
            {
                for (i = 0; i < l; i += 2) sum2 += a[i];
                for (i = l + 1; i <= r; i += 2) sum2 += a[i];
                for(i=r+1;i<N;i+=2) sum2+=a[i];
            }

            sum=Math.max(sum,sum2);

            if(r2!=-1&&l2!=-1)
            {
                for(i=0;i<l2;i+=2) sum3+=a[i];
                for(i=l2;i<r2;i+=2) sum3+=a[i];

                r2++;
                if(r2%2==1) r2++;
                for(i=r2;i<N;i+=2) sum3+=a[i];
            }
            sum=Math.max(sum,sum3);

            sb.append(sum).append("\n");
        }
        System.out.println(sb);
    }
}