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 stdout, stdin


def maxSubArraySum(a):
	size = len(a)
	if size == 0:
		return 0
	max_so_far = a[0]
	curr_max = a[0]

	for i in range(1, size):
		curr_max = max(a[i], curr_max + a[i])
		max_so_far = max(max_so_far, curr_max)

	return max_so_far


for _ in range(int(stdin.readline())):
	n = int(stdin.readline())
	ls = list(map(int, stdin.readline().split()))
	if n % 2 == 1:
		ls.append(0)
	n += 1
	sm = 0
	diff = []
	diff2 = []
	for i in range(0, n - 1, 2):
		sm += ls[i]
		diff.append(ls[i + 1] - ls[i])
	for i in range(1, n - 2, 2):
		diff2.append(ls[i] - ls[i + 1])
	a = maxSubArraySum(diff)
	b = maxSubArraySum(diff2)
	stdout.write(str(max(max(a, b), 0) + sm) + '\n')
