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
input=sys.stdin.readline
def kadane(l):
    max_so_far = 0
    max_ending_here = 0
    for i in range(len(l)): 
        max_ending_here = max_ending_here + l[i] 
        if max_ending_here < 0: 
            max_ending_here = 0
        elif (max_so_far < max_ending_here): 
            max_so_far = max_ending_here 
    return(max_so_far) 
t=int(input())
for ii in range(t):
    n=int(input())
    a=[int(i) for i in input().split()]
    if n==1:
        print(a[0])
    elif n==2:
        print(max(a[1],a[0]))
    else:
        evensum=0
        for i in range(0,n,2):
            evensum+=a[i]
        freq1,freq2=[],[]
        freq1.append(max(0,a[1]-a[0]))
        freq2.append(max(0,a[1]-a[2]))
        for i in range(3,n,2):
            
                freq1.append((a[i]-a[i-1]))
        if not n&1:
            n=n-1
        for i in range(3,n,2):
            
                freq2.append((a[i]-a[i+1]))
        p,q=kadane(freq1),kadane(freq2)
        #print(evensum,freq1,freq2)
        print(evensum+max(p,q))
                       
    
