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
# from functools import lru_cache
from sys import stdin, stdout
import sys
from math import *
# from collections import deque
# sys.setrecursionlimit(3*int(1e5))
# input = stdin.buffer.readline
# print = stdout.write
 
# @lru_cache()

def kedane(a,n):
	ans=0
	curr=0
	for i in range(n):
		curr+=a[i]
		if(curr>ans):
			ans=max(ans,curr)
		elif(curr<=0):
			curr=0
	return ans


for __ in range(int(input())):
	n=int(input())
	ar=list(map(int,input().split()))
	a=[]
	for i in range(1,n):
		if(i%2==1):
			a.append(ar[i]-ar[i-1])
	b=[]
	for i in range(2,n):
		if(i%2==0):
			b.append(ar[i-1]-ar[i])

	ans=0
	for i in range(n):
		if(i%2==0):
			ans+=ar[i]
	
	print(max(ans,ans+kedane(a,len(a)),ans+kedane(b,len(b))))






