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
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
from copy import *
from bisect import *	
from heapq import *
from math import *
from itertools import permutations 
 
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
 
#sys.setrecursionlimit(10**6)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
def li():
	return [int(x) for x in input().split()]
def fi():
	return int(input())
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())	
def gh():
	sys.stdout.flush()
def graph(n,m):
	for i in range(m):
		x,y=mi()
		a[x].append(y)
		a[y].append(x)
def bo(i):
	return ord(i)-ord('a')

def cal(s):
	res = 0
	for init in range(1000):
	    cur = init
	    ok = True
	    for i in range( len(s)):
	        res = res + 1
	        if s[i] == '+':
	            cur = cur + 1
	        else:
	            cur = cur - 1
	        #print(init,cur)    
	        if cur < 0:
	            ok = False
	            break

	    if ok:
	    	#print("CUR",cur)
	    	break
	print(res)        

t=fi()
def sd(a):
	c=0
	for i in range(0,n,2):
		c+=a[i]
	return c
#POINTS: After r or before l nothing will be changed
#If we take subarray of odd length it does not help
#So choose the subarray such that even-odd is maximum
#How to do that?	
#The elements in the subarray parity will same as original array
def ka(a):
	maxi=0
	c=0
	for i in range(len(a)):
		
		if c+a[i]>a[i]:
			c+=a[i]
		else:
			c=a[i]
		maxi=max(maxi,c)
	#print(a,maxi)		
	return maxi		
while t>0:
	t-=1
	n=fi()
	a=li()
	maxi=sd(a)
	"""for i in range(n):
		for j in range(i+1,n+1):
			b=deepcopy(a)
			b=b[:i]+b[i:j][::-1]+b[j:]
			if sd(b)>maxi:
				maxi=sd(b)
				lol=deepcopy(b[i:j])"""
	d=[]
	e=[]			
	for i in range(0,n,2):
		if len(d):
			d.append(d[-1]+a[i])
		else:
			d.append(a[i])
	for i in range(1,n,2):
		if len(e):
			e.append(e[-1]+a[i])
		else:
			e.append(a[i])
	l1=[]
	l2=[]
	for i in range(1,n,2):
		l1.append(a[i]-a[i-1])	
	for i in range(2,n,2):
		l2.append(-a[i]+a[i-1])		
	print(sd(a)+max(ka(l1),ka(l2)))