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 kadene(A):
    max_g=0
    max_c=0
    for i in range(len(A)):
        max_c=max(A[i], max_c+A[i])
        if max_c>max_g:
            max_g=max_c
    return max_g
        
        
        
def answer(n,A):
    if n==1:
        return A[0]
    dp=[0]*(n//2)
    count=0
    s=0
    for i in range(1,n,2):
        dp[count]=A[i]-A[i-1]
        count+=1
        
        s+=A[i-1]
        
    dp1=[0]*(n//2)
    count=0
    for i in range(1,n,2):
        if i+1<n:
            dp1[count]=A[i]-A[i+1]
            count+=1

    k1=kadene(dp)
    k2=kadene(dp1)
    if n%2!=0:
        s+=A[-1]
    
    return s+max(k1,k2)
    
t=int(input())
for i in range(t):
    n=int(input())
    A=list(map(int,input().split()))
    print(answer(n,A))