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
t = int(input())
for _ in range(t) :
    n = int(input())
    line = input()
    ar = line.split()
    ar = [int(i) for i in ar]
    ans1 = list()
    ans2 = [0, 0]
    ans3 = list()
    ans1.append(0)
    ans3.append(0)
    for i in range(n) :
        if i % 2 == 0 :
            ans1.append(ans1[i] + ar[i])
        else :
            ans1.append(ans1[i] + 0)
        if i % 2 == 0 and i+2 <= n:
            ans2.append(ar[i+1] + max(ans1[i], ans2[i]))
        elif i+2 <= n :
            ans2.append(ar[i] + max(ans1[i], ans2[i]))
        if i % 2 == 0 :
            ans3.append(max(ans1[i], ans2[i], ans3[i]) + ar[i])
        else :
            ans3.append(max(ans1[i], ans2[i], ans3[i]) + 0)
    print(max(max(ans1), max(ans2), max(ans3)))
            
    