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.Arrays;
import java.util.Scanner;

public class er90d {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int t=scan.nextInt();
		for(int tt=0;tt<t;tt++) {
			int n=scan.nextInt();
			long[] a=new long[n];
			long init=0L;
			for(int i=0;i<n;i++) {
				a[i]=scan.nextInt();
				if(i%2==0) init+=a[i];
			}
			long[] sum=new long[(n+1)/2];
			long res=0L;

			for(int i=0;i+1<n;i+=2) {
				sum[i/2]=a[i+1]-a[i];
			}
			res=Math.max(res,init+maxSubarray(sum));

			Arrays.fill(sum,0);
			for(int i=1;i+1<n;i+=2) {
				sum[i/2]=a[i]-a[i+1];
			}
			res=Math.max(res,init+maxSubarray(sum));
			System.out.println(res);
		}
	}
	public static long maxSubarray(long[] a) {
		long sum=0L, res=0L;
		for(int i=0;i<a.length;i++) {
			sum+=a[i];
			sum=Math.max(sum,0);
			res=Math.max(sum,res);
		}
		return res;
	}
}