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 os
import sys
input=sys.stdin.buffer.readline
from math import*

t=int(input())
while t:
	t-=1
	n=int(input())
	arr=list(map(int,input().split()))
	if n==1:
		print(sum(arr))
	else:
		brr=[]
		crr=[]
		s=0
		i=0
		while i<n:
			s+=arr[i]
			i+=2
		i=0
		while i<n:
			if i+1<n:
				brr.append(arr[i+1]-arr[i])
			if i>0:
				crr.append(arr[i-1]-arr[i])
			i+=2
		maxbrr=0
		maxcrr=0
		c=0
		for x in brr:
			c+=x
			if c<0:
				c=0
			maxbrr=max(maxbrr,c)
		c=0
		for x in crr:
			c+=x
			if c<0:
				c=0
			maxcrr=max(maxcrr,c)
#		print(brr)
#		print(crr)
		print(s+max(maxbrr,maxcrr))
