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
t=int(input())
import random
xx=random.randint

for _ in range(t):
    n=int(input())
    l=list(map(int,input().split()))
    
    if n%2:
        l+=[0]*3
        n+=3
    else:
        l+=[0]*2
        n+=2
    odd=[]
    even=[0]
    for i in range(n-1,-1,-1):
        if i%2:
            if odd:
                #print(even,odd)
                y = l[i]+max(even[-1],odd[-1][1])
                x=0
                if len(odd)-1:
                    x = l[i]+max(odd[-1][0],even[-2])
                odd.append([x,y])
            else:
                
                odd.append([0,l[i]])
        else:
            even.append(even[-1]+l[i])
    odd.reverse()
    ans=odd[0][1]
    even=0
    b=0
    for i in range(n):
        if i%2:
            b+=1
        else:
            even+=l[i]
            if b+1<len(odd):
                ans=max(ans,even+odd[b+1][1])
            
            ans=max(ans,even+odd[b][0])
    print(ans)
            
        
        
    
            
    
        
