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
nb_cases = int(input())
for case in range(nb_cases):
    n = int(input())
    a = [int(x) for x in input().split()]
    dp_pair = []
    dp_impair = []
    for x in range(n-1):
        if x%2==0:
            dp_pair.append(a[x+1] - a[x])
        else:
            dp_impair.append(a[x] - a[x+1])
    # pair
    p1 = 0
    p2 = 0
    somme = 0
    p_min = -1
    p_max = -1
    somme_max = 0
    while p2 < int(n/2):
        somme += dp_pair[p2]
        if dp_pair[p2] > 0:
            if somme > somme_max:
                somme_max = somme
                p_min = p1
                p_max = p2
        else:
            if somme <= 0:
                p1 = p2+1
                somme = 0
        p2 += 1
    somme_max_pair = somme_max

    # impair
    p1 = 0
    p2 = 0
    somme = 0
    p_min = -1
    p_max = -1
    somme_max = 0
    while p2 < int((n-1)/2):
        somme += dp_impair[p2]
        if dp_impair[p2] > 0:
            if somme > somme_max:
                somme_max = somme
                p_min = p1
                p_max = p2
        else:
            if somme <= 0:
                p1 = p2+1
                somme = 0
        p2 += 1
    somme_max_impair = somme_max
    somme_pair = sum([x for x in a[::2]])
    print(somme_pair + max(somme_max_pair, somme_max_impair))
