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 kadane_sum(lista): #Algorytm Kadane = zwraca największa suma podciągu
    max_local_np = max_local_p = max_local_np2 = max_local_p2 = 0
    best, best2= 0, 0
    for y in range(1, len(lista), 2):
        max_local_p += lista[y - 1]
        max_local_np += lista[y]
        best = max (best, max_local_np - max_local_p)
        if max_local_p > max_local_np:
    	    best = max(best, max_local_np - max_local_p)
    	    max_local_np = 0
    	    max_local_p = 0
        best = max(best, max_local_np - max_local_p)
    for y in range(2, len(lista), 2):
        max_local_p2 += lista[y]
        max_local_np2 += lista[y - 1]
        best2 = max (best2, max_local_np2 - max_local_p2)
        if max_local_p2 > max_local_np2:
    	    best2 = max(best2, max_local_np2 - max_local_p2)
    	    max_local_np2 = 0
    	    max_local_p2 = 0
        best2 = max(best2, max_local_np2 - max_local_p2)
    best = max(best, best2)
    return best
for _ in range (int(input())):
    ilosc = input()
    lista = [ int(x) for x in input().split()]
    best = kadane_sum(lista)
    print(sum(lista[::2]) + best)
