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, math
input = sys.stdin.readline

def getInts():
    return [int(s) for s in input().split()]

def getInt():
    return int(input())

def getStrs():
    return [s for s in input().split()]

def getStr():
    return input()

def listStr():
    return list(input())

import collections as col
import math

def solve():
    N = getInt()
    A = getInts()
    if N == 1:
        return A[0]
    if N == 2:
        return (max(A[0],A[1]))
    #we can either swap an array starting on even, ending on odd, or vice versa
    #take the pairwise differences in each case (a0,a1) (a2,a3) etc and (a1,a2) (a3,a4) etc
    #we then want the greatest positive sum amongst them
    #so we have D1 D2 D3 ... D(N//2)
    #and then we have E1 E2 E3 ... E(N//2)
    #we the answer is the initial sum of even positions, much the maximal contiguous subarray sum from either the Ds or the Es
    D = []
    E = []
    evens = sum([A[j] for j in range(0,N,2)])
    for j in range(0,N-1,2):
        D.append(A[j+1]-A[j])
    for j in range(1,N-1,2):
        E.append(A[j]-A[j+1])
    #if at any point the sum goes negative, we skip
    curr_best = 0
    D_ind = 0
    curr_sum = 0
    while D_ind < len(D):
        curr_sum += D[D_ind]
        curr_best = max(curr_sum,curr_best)
        if curr_sum < 0:
            curr_sum = 0
        D_ind += 1
    E_ind = 0
    curr_sum = 0
    while E_ind < len(E):
        curr_sum += E[E_ind]
        curr_best = max(curr_sum,curr_best)
        if curr_sum < 0:
            curr_sum = 0
        E_ind += 1
    return evens + curr_best

    
for _ in range(getInt()):
    print(solve())
    

