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
# https://codeforces.com/contest/1373/problem/D
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
# do magic here


def kadane(A):
    # print(A)
    t = -100000000000
    s = 0
    for x in A:
        t = max(x, t+x)
        s = max(s, t)
    return s


t = int(input())
for _ in range(t):
    n = int(input())
    arr = [int(x) for x in input().split()]
    diff1 = [arr[x]-arr[x-1] for x in range(1, n, 2)]
    diff2 = [arr[x-1]-arr[x] for x in range(2, n, 2)]
    ans = sum([arr[x] for x in range(0, n, 2)])
    temp1 = kadane(diff1)
    temp2 = kadane(diff2)
    temp = max(temp1,temp2)
    if temp > 0:
        ans += temp
    print(ans)
