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.Scanner;

public class Main {
    static long aa[][][];
    static long maxx(int ind,int flag,long[] arr,int flag1){
        if(ind>=arr.length && flag1==0){
            return 0;
        }
        if(ind>=arr.length && flag1==1){
            return Integer.MIN_VALUE;
        }
        if(aa[ind][flag][flag1]!=-1){
            return aa[ind][flag][flag1];
        }
        if(flag==0){
            if(ind%2==0) {
                return aa[ind][flag][flag1]= Math.max(maxx(ind + 1, 0, arr, flag1) + arr[ind], maxx(ind + 1, 1, arr, 1));
            }
            else{
                return aa[ind][flag][flag1]= Math.max(maxx(ind+1,0,arr,flag1),maxx(ind+1,1,arr,1)+arr[ind] );
            }
        }
        else if(flag==1){
            if(flag1==1){
                if(ind%2==0){
                    return aa[ind][flag][flag1]= maxx(ind+1,1,arr,0);
                }
                else{
                    return aa[ind][flag][flag1]= maxx(ind+1,1,arr,0)+arr[ind];
                }
            }
            else{
                if(ind%2==0){
                    return aa[ind][flag][flag1]= Math.max(maxx(ind+1,2,arr,0)+arr[ind],maxx(ind+1,1,arr,1));
                }else{
                    return aa[ind][flag][flag1]= Math.max(maxx(ind+1,2,arr,0),maxx(ind+1,1,arr,1)+arr[ind]);
                }
            }
        }
        else{
            if(ind%2==0){
                return aa[ind][flag][flag1]= maxx(ind+1,flag,arr,flag1)+arr[ind];
            }
            else{
                return aa[ind][flag][flag1]= maxx(ind+1,flag,arr,flag1);
            }
        }

    }

    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int t=scanner.nextInt();
        for(int j=0;j<t;j++){
            int n=scanner.nextInt();
            long[] arr=new long[n];
            for(int i=0;i<n;i++){
                arr[i]= scanner.nextLong();
            }
            aa=new long[n][3][2];
            for(int i=0;i<n;i++){
                for(int k=0;k<3;k++){
                    for(int l=0;l<2;l++){
                        aa[i][k][l]=-1;
                    }
                }
            }
            System.out.println(maxx(0,0,arr,0));

        }
    }
}
