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
from sys import stdin

for _ in range(int(stdin.readline())):
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().split()))
    res = 0
    max_so_far_before = 0
    max_ending_before = 0
    max_so_far_after = 0
    max_ending_after = 0
    for i, v in enumerate(a):
        if i % 2 == 0:
            res += v
        else:
            max_ending_before += v - a[i-1]
            if max_ending_before < 0:
                max_ending_before = 0
            if max_so_far_before < max_ending_before:
                max_so_far_before = max_ending_before
            if i < n - 1:
                max_ending_after += v - a[i+1]
                if max_ending_after < 0:
                    max_ending_after = 0
                if max_so_far_after < max_ending_after:
                    max_so_far_after = max_ending_after
    res += max(max_so_far_after, max_so_far_before)
    print(res)