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
## wee need to find the even length sub array such that the 
##val sum(even_indexes) - sum(odd_indexes) is maximized

def kadane_alg(a):
    max_so_far=float("-inf")
    max_ending_here=0
    
    for i in range(len(a)):
        max_ending_here=max_ending_here+a[i]
        if max_so_far < max_ending_here:
            max_so_far=max_ending_here
        
        if max_ending_here < 0:
            max_ending_here = 0
    
    return max_so_far if max_so_far > 0 else 0

for _ in range(int(input())):
    n=int(input())
    arr=[int(c) for c in input().split()]
    i=0
    newarr=[0 for i in range(len(arr)//2)]
    j=0
    while i+1 < len(arr):
        newarr[j] = arr[i+1] - arr[i]
        i+=2
        j+=1
    i=1
    newarr2=[0 for i in range(len(arr)//2)]
    j=0
    while i+1 < len(arr):
        newarr2[j] = arr[i] - arr[i+1]
        i+=2
        j+=1
    
    evensum=0
    for i in range(0,len(arr),2):
        evensum+=arr[i]
    
    print(evensum+max(kadane_alg(newarr),kadane_alg(newarr2)))
    
    
    