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 sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7

def ni():
    return int(raw_input())


def li():
    return map(int,raw_input().split())


def pn(n):
    stdout.write(str(n)+'\n')


def pa(arr):
    pr(' '.join(map(str,arr))+'\n')

# fast read function for total integer input

def inp():
    # this function returns whole input of
    # space/line seperated integers
    # Use Ctrl+D to flush stdin.
    return map(int,stdin.read().split())

range = xrange # not for python 3.0+

for t in range(ni()):
    n=ni()
    l=li()
    ans=0
    for i in range(0,n,2):
        ans+=l[i]
    sm1=0
    a1=0
    for i in range(0,n-1,2):
        val=l[i+1]-l[i]
        #print val
        sm1+=val
        if sm1<0:
            sm1=0
            continue
        #sm1+=val
        a1=max(sm1,a1)
    sm2=0
    a2=0
    for i in range(1,n-1,2):
        val=l[i]-l[i+1]
        sm2+=val
        #print val
        if sm2<0:
            sm2=0
            continue
        #sm2+=val
        a2=max(a2,sm2)
    pn(ans+max(a1,a2))
