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
for k in range(0,int(input())):
    a = int(input())
    l1 = list(map(int,input().split()))
    l2 = []
    l3 = []
    A = 0 
    B = 0
    C = 0
    D = 0
    s1 = 0
    for u in range(0,len(l1),2):
        s1+=l1[u]
    for i in range(0,len(l1)-1,2):
        l2.append(l1[i+1] - l1[i])
    for j in range(1,len(l1)-1,2):
        l3.append(l1[j] - l1[j+1])
    for l in l2:
        A+=l
        if A < 0:
            A = 0
        if A>B:
            B = A
    for g in l3:
        C+=g
        if C < 0:
            C = 0        
        if C>D:
            D = C
    if B>D:
        print(s1+B)
    else:
        print(s1+D)