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
#!/usr/bin/env python3
import io
import os

input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

def get_str():
    return input().decode().strip()

def rint():
    return map(int, input().split())

def oint():
    return int(input())

def find_max(diff):
    max_sum = 0
    sum_ = 0
    for i in range(len(diff)):
        sum_ += diff[i]
        if sum_ < 0:
            sum_ = 0
        else:
            max_sum = max(max_sum, sum_)
    return max_sum

t = oint()

for _ in range(t):
    n = oint()
    a = list(rint())
    diff_even = []
    for i in range(0, n-1, 2):
        diff_even.append(a[i+1]- a[i])
    diff_odd = []
    for i in range(1, n-1, 2):
        diff_odd.append(a[i]- a[i+1])
    ans = sum(a[::2]) + max(find_max(diff_even), find_max(diff_odd))
    print(ans)
