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 sys

def kadane(b,ch):

	mx=b[0]
	cur=0

	ms=0
	me=0
	start=0

	if(ch==1):
		ms+=1
		me+=1

	for h in range(len(b)):
		if(cur+b[h]>=0):
			if(cur<0):
				cur=b[h]
				start=h
			else:
				cur+=b[h]
		else:
			if(cur>mx):
				mx=cur
				ms=2*start
				me=2*h

				if(ch==1):
					ms+=1
					me+=1
			
			cur=b[h]
			start=h

		if(cur>mx):
			mx=cur
			ms=2*start
			me=2*h

			if(ch==1):
				ms+=1
				me+=1

	if(cur>mx):
		mx=cur
		ms=2*start
		me=2*(len(b)-1)

		if(ch==1):
			ms+=1
			me+=1

	return mx,ms,me

def reverse(arr,i,j):
	
	for p in range(i,j+1,2):
		arr[p],arr[p+1]=arr[p+1],arr[p]

t=int(sys.stdin.readline())

ans_arr=[]

for i in range(t):
	n=int(sys.stdin.readline())

	a=[int(x) for x in sys.stdin.readline().split()]

	if(n==1):
		ans_arr.append(str(a[0]))
	elif(n==2):
		ans_arr.append(str(max(a[0],a[1])))
	else:

		ans=0

		for e in range(n):
			if(e%2==0):
				ans+=a[e]

		b1=[]
		b2=[]

		for g in range(0,n,2):
			if(g+1<n):
				b1.append(a[g+1]-a[g])

		for h in range(1,n,2):
			if(h+1<n):
				b2.append(a[h]-a[h+1])

		mx1,ms1,me1 = kadane(b1,0)
		mx2,ms2,me2 = kadane(b2,1)

		ans1=0
		ans2=0

		if(mx1>mx2):
			ms=ms1
			me=me1
			reverse(a,ms,me)

			for p in range(len(a)):
				if(p%2==0):
					ans1+=a[p]
		else:
			ms=ms2
			me=me2

			reverse(a,ms,me)

			for p in range(len(a)):
				if(p%2==0):
					ans2+=a[p]

		ans=max(ans,ans1)
		ans=max(ans,ans2)	

		ans_arr.append(str(ans))

print("\n".join(ans_arr))
