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
t=int(input())

def f(a):
	sums=[0]
	for i in range(len(a)):
		sums.append(a[i]+sums[-1])
	minn=[sums[0]]
	for i in range(1,len(sums)):
		minn.append(min(minn[-1],sums[i]))
	maxx=[sums[-1]]
	for i in range(1,len(sums)):
		maxx.append(max(maxx[-1],sums[len(sums)-i-1]))
	maxx.reverse()
	diff=[0]
	for i in range(1,len(sums)):
		diff.append(maxx[i]-minn[i])
	return(max(diff))




for tt in range(t):
	n=int(input())
	a=[int(x) for x in input().split(' ')]
	ch=[]
	for i in range(1,n):
		if(i%2==1):
			ch.append(a[i]-a[i-1])
		else:
			ch.append(a[i-1]-a[i])
	# print(ch)
	l=[]
	r=[]
	for i in range(len(ch)):
		if(i%2==0):
			l.append(ch[i])
		else:
			r.append(ch[i])
	# print(l)
	# print(r)
	sum_org=0
	for i in range(n):
		if(i%2==0):
			sum_org+=a[i]
	# print(sum_org)
	summ=sum_org+max(f(l),f(r),0)
	print(summ)