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
for _ in range(int(input())):
    n = int(input())
    lis = list(map(int,input().split()))
    eve=[]
    odd=[]
    for i in range(1,n,2):
        eve.append(lis[i]-lis[i-1])
    for i in range(2,n,2):
        odd.append(lis[i]-lis[i-1])
    l=len(eve);i=j=0
    l1=len(odd)
    while i<l and eve[i]<=0:
        i+=1
    while j<l1 and odd[j]>=0:
        j+=1    
    ans1=ans2=tmp=0
    while i<l:
        if eve[i]>=0:
            tmp+=eve[i]
        else:
            ans1=max(ans1,tmp)
            if tmp+eve[i]<0:
                tmp=0
            else:
                tmp+=eve[i]
        i+=1
    ans1=max(ans1,tmp)
    tmp=0  
    while j<l1:
        if odd[j]<=0:
            tmp+=odd[j]
        else:
            ans2=min(ans2,tmp)
            if tmp+odd[j]>0:
                tmp=0
            else:
                tmp+=odd[j]
        j+=1
    ans2=abs(min(ans2,tmp))
#    print(ans1,eve,odd,ans2)
    ans=0
    for i in range(0,n,2):
        if i%2==0:
            ans+=lis[i]     
    print(max(ans1,ans2)+ans)                              
