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
# O(n)
import math

def solution(n, arr):
    if n == 1:
        return arr[0]
    sume = 0
    sumarr = [] #the sumarray
    for i in range(n):
        sume += (-1)**(i+1)*arr[i]
        sumarr.append(sume)
    
    def sum(i): # comodin function sum
        if i < 0: return 0
        return -math.inf if i >= n else sumarr[i]
    
    maxims = list(range(n)) #searching the maxims
    maxims[-1], maxims[-2] = n, n-1
    for i in range(n - 3, -1, -1):
        if sum(i+1) >= sum(maxims[i+2]):
            maxims[i] = i+1
        else:
            maxims[i] = maxims[i+2]
    
    maximum = 0 #searching the maximum subarray
    for i in range(n):
        if sum(maxims[i]) - sum(i-1) > maximum:
            maximum = sum(maxims[i]) - sum(i-1)
    
    sume = 0 #final sum
    for i in range(0, n, 2):
        sume += arr[i]    
    return sume + maximum

cases = int(input())
for case in range(cases):
    print(solution(int(input()), [int(x) for x in input().split(' ')]))
