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
/*

Solution:

Reversing an odd subarray does nothing

We reverse an even subarray. We use PREFIX sums on the CHANGE OF SUM.
Let psa[i]= psa[i-1] - (-1)^ia[i] (psa[0]=-a[0])
We simply find the greatest psa[i] and the smallest psa[j], 
while making sure i,j have different parities.
*/
import java.io.*;
import java.util.*;
public class CF1373D {
    public static void main(String[] args) throws IOException{
        
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int T=Integer.parseInt(br.readLine());
        int N; long[] psa; long sum; int cur;
        for (int i = 0; i < T; i++) {
            N=Integer.parseInt(br.readLine());
            psa=new long[N+1]; //psa[0]=0;
            sum=0; // Original sum of even positions
            StringTokenizer st=new StringTokenizer(br.readLine());
            cur=Integer.parseInt(st.nextToken()); sum+=cur;
            psa[1]=-cur;
            for (int j = 2; j <= N; j++) {
                cur=Integer.parseInt(st.nextToken()); 
                if(j%2==1){
                    sum+=cur;
                    psa[j]=psa[j-1]-cur;
                    
                }else{
                    psa[j]=psa[j-1]+cur;
                    
                }
            }
            
            //Now consider max (psa[i]-psa[j]), i>j, 2 divides i-j
            //This is equivalent to reversing subarray [j+1,i]
            //To solve this, we traverse the array psa. 
            //Let min1 be the minimal element among the traversed elements with odd indices
            //Define min2 similarly
            //Keep on taking the max of (cur-min), updating min
            
            long min2=0; //min2 is at most 0, psa[0]=0;
            long min1=psa[1];
            long max=0; //Change
            for (int j = 2; j <=N; j++) {
                if(j%2==0){
                    if(psa[j]<min2){
                        min2=psa[j];
                    }
                    max=Math.max(psa[j]-min2,max);
                }else{
                    if(psa[j]<min1){
                        min1=psa[j];
                    }
                    max=Math.max(psa[j]-min1,max);
                }
                
            }
            System.out.println(sum+max);
        }
    }
}
