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
def kadane(l):
    length=len(l)
    maxsub=globalmax=l[0]
    for i in range(1,length):
        maxsub=max(l[i],maxsub+l[i])
        globalmax=max(globalmax,maxsub)
    return globalmax





t=int(input(""))
while t>0:
    n=int(input(""))
    a=list(map(int,input().split()))
    neighbours=[]

    oddsum=evensum=0
    l1=[0]
    l2=[0]
    
    #print(a)
    if(n==1):
        ans=a[0]
    else:
        for i in range(0,len(a),2):
            if(i!=len(a)-1):
                l1.append(a[i+1]-a[i])
            evensum+=a[i]
        ans=evensum
        #print("evensum",evensum)
        for j in range(1,len(a),2):
            if(j!=len(a)-1):
                l2.append(a[j]-a[j+1])
        #print("l1 and l2",l1,l2)
        change=max(kadane(l1),kadane(l2))
        if(change>0):
            ans+=change
    print(ans)       
        
        
    t-=1