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
def f(al):
    ar1=[]
    ar2=[]
    i=0
    while i<len(al)-1:
        ar1.append(al[i+1]-al[i])
        i=i+2
    i=0
    while i<len(al)-2:
        ar2.append(al[i+1]-al[i+2])
        i=i+2
    return (ar1,ar2)
    
def subarraymax(nums):
    if len(nums)==0:
        return 0
    sum = res = nums[0]
    for i in range(1,len(nums)):
        sum = max(nums[i], sum+nums[i])
        res = max(res, sum)
    return res
    
for tc in range(int(input())):
    _=input()
    al=list(map(int,input().split()))
    ini=0
    for i in range(0,len(al),2):
        ini=ini+al[i]
    s1,s2=f(al)
    delta=max(subarraymax(s1),subarraymax(s2))
    if delta<=0:
        print(ini)
    else:
        print(ini+delta)