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 math import ceil
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    summ = 0
    ch0 = 0
    ch1 = 0
    for i in range(len(a)):
        if i % 2 == 0:
            ch0 += a[i]
    v1 = [0 for i in range(n // 2)]
    v2 = [0 for i in range((n - 1) // 2)]
    for i in range(n // 2):
        v1[i] = a[2 * i + 1] - a[2 * i]
    for i in range(0, (n - 1) // 2):
        v2[i] = a[2 * i + 1] - a[2 * i + 2]
    csum = 0
    msum = 0
    ans1 = ch0
    for i in v1:
        csum += i
        msum = min(csum, msum)
        ans1 = max(csum - msum + ch0, ans1)
    csum = 0
    msum = 0
    ans2 = ch0
    for i in v2:
        csum += i
        msum = min(csum, msum)
        ans2 = max(csum - msum + ch0, ans2)
    print(max(ans1, ans2))