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 os,io
from sys import stdout
import collections
import random
import math
from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import Counter
import heapq

# import sys
# sys.setrecursionlimit(10**6)

# from functools import lru_cache
# @lru_cache(maxsize=None)

def primes(n):
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
    return [2] + [i for i in range(3,n,2) if sieve[i]]

def binomial_coefficient(n, k):
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in range(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

def powerOfK(k, max):
    if k == 1:
        return [1]
    if k == -1:
        return [-1, 1]

    result = []
    n = 1
    while n <= max:
        result.append(n)
        n *= k
    return result

def prefixSum(arr):
    for i in range(1, len(arr)):
        arr[i] = arr[i] + arr[i-1]
    return arr

def divisors(n):
    i = 1
    result = []
    while i*i <= n:
        if n%i == 0:
            if n/i == i:
                result.append(i)
            else:
                result.append(i)
                result.append(n/i)
        i+=1
    return result

def kadane(a,size):
    max_so_far = 0
    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

# s = input().decode('utf-8').strip()
t = int(input())
for _ in range(t):
    n = int(input())
    l = list(map(int, input().split()))

    left = []
    for i in range(0, len(l)-1, 2):
        left.append(l[i+1]-l[i])
    right = []
    for i in range(1, len(l)-1, 2):
        right.append(l[i]-l[i+1])

    a = kadane(left, len(left))
    b = kadane(right, len(right))
    print(sum(l[::2]) + max(a, b))
