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
# for _ in  range(int(input())):
# 	n,x = map(int,input().split(" "))
# 	arr = list(map(int,input().split(" ")))

# 	x = 1-(x/100)

# 	start = 0
# 	end = 1000

# 	want = 0
# 	give = 1
# 	curr = 0

# 	while abs(want - give) >= 1e-7 :
# 		curr = (start + end)/2

# 		want = 0
# 		give = 0
# 		for i in arr:
# 			want += max(0,curr - i)
# 		for i in arr:
# 			give += max(0,i - curr)*x

# 		if(want>give):
# 			end = curr
# 		else:
# 			start = curr		
# 	print(curr)

# import heapq

# n,k = map(int,input().split(" "))
# arr = list(map(int,input().split(" ")))

# arr1 = []
# for i in range(n):
# 	arr1.append([arr[i],i])

# heapq.heapify(arr1)
# for i in range(k):
# 	popped = heapq.heappop(arr1)
# 	popped[0]+=1
# 	heapq.heappush(arr1,popped)
# arr1.sort(key = lambda x:x[1])
# for i in range(n):
# 	print(arr1[i][0],end= " ")
# print()

# dp = []
# for i in range(1000):
# 	for j in range(1000):
# 		dp.append(i**2+j**2)
# dp = list(set(dp))
# dp.sort()

# for _ in range(int(input())):
# 	print(dp[int(input())-1])

from sys import stdin

def maxSubArraySum(a,size): 
	max_so_far = -100000000000
	max_ending_here = 0
	for i in range(0, size): 
		max_ending_here = max_ending_here + a[i] 
		if (max_so_far < max_ending_here): 
			max_so_far = max_ending_here 
		if max_ending_here < 0: 
			max_ending_here = 0   
	return max_so_far 

def help():
	n = int(stdin.readline())
	arr = list(map(int,stdin.readline().split(" ")))

	ans = 0

	for i in range(0,n,2):
		ans += arr[i]

	arr1 = []
	arr2 = []

	for i in range(1,n-1,2):
		arr1.append(arr[i]-arr[i+1])

	for i in range(0,n-1,2):
		arr2.append(arr[i+1]-arr[i])
	# print(arr1,arr2)

	tt = max(maxSubArraySum(arr1,len(arr1)),maxSubArraySum(arr2,len(arr2)))
	# print(ans,tt)
	tt = max(0,tt)

	print(ans+tt)


for _ in range(int(stdin.readline())):
	help()