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 helper(A):
    MIN = []
    MAX = []
    temp_min = max(A)+1
    for i in range(len(A)):
        temp_min = min(temp_min,A[i])
        MIN += [temp_min]
    
    temp_max = min(A)-1
    A = A[::-1]
    for i in range(len(A)):
        temp_max = max(temp_max,A[i])
        MAX += [temp_max]
    MAX = MAX[::-1]
    ans = A[0] - A[-1]
    #print(MIN,MAX)
    for i in range(len(A)-1):
        ans = max(ans,MAX[i+1]-MIN[i])
    return ans

def find(A):
    pre_even = []
    pre_odd = []
    temp_even = 0
    temp_odd = 0
    for i in range(len(A)):
        if i%2==0:
            temp_even += A[i]
        else:
            temp_odd += A[i]
        pre_even += [temp_even]
        pre_odd += [temp_odd]
    # pre_even[i] = \sum_{k=0}^{i}_{k%2==0} A[i]
    # L,R 
    # new_total = total - (pre_even[R]-pre-even[L-1])+(pre_odd[R]-pre_odd[L-1])
    #           += (pre_odd[R]-pre_even[R])-(pre_odd[L-1]-pre_even[L-1])
    ans_odd = [0]
    ans_even = []
    for i in range(len(A)):
        if i%2==0:
            ans_even+=[pre_odd[i]-pre_even[i]]
        else:
            ans_odd+=[pre_odd[i]-pre_even[i]]
    #print(ans_even,ans_odd)
    ans1 = helper(ans_odd)
    ans2 = helper(ans_even)
    #print(pre_even[-1],ans1,ans2)
    return max(pre_even[-1]+ans1,pre_even[-1]+ans2,pre_even[-1])
    
    # given A, find max_{i<j}(A[j]-A[i])
for _ in range(int(input())):
    input()
    A = list(map(int,input().strip().split()))
    print(find(A))
    