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
import sys
import math

read = sys.stdin.buffer.read
# input = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
t = int(input())
for case in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    """step1 : 現在の偶数番目の和を計算
step2: 累積和を計算する
step3: 偶数番目、奇数番目それぞれ累積和の列に対して、現在の最大、最小をそれぞれ更新して差分のmaxを計算
"""
    origin = sum([a[i] for i in range(n) if i % 2 == 0])
    acc = [0] * (1 + n)
    for i in range(n):
        if i % 2:
            acc[i + 1] = acc[i] + a[i]
        else:
            acc[i + 1] = acc[i] - a[i]
    # 偶数に対して現在のmax,minを更新=> min のみの管理
    ans_dif = 0
    # print(origin, acc)
    even_min = acc[0]
    for i in range(n // 2 + 1):
        even_min = min(even_min, acc[i * 2])
        ans_dif = max(ans_dif, acc[i * 2] - even_min)

    odd_min = acc[1]
    for i in range((1 + n) // 2):
        odd_min = min(odd_min, acc[i * 2 + 1])
        ans_dif = max(ans_dif, acc[i * 2 + 1] - odd_min)
    print(ans_dif + origin)

