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
FAST_IO = 0
if FAST_IO:
    import io, sys, atexit
    rr = iter(sys.stdin.read().splitlines()).next
    sys.stdout = _OUTPUT_BUFFER = io.BytesIO()
    @atexit.register
    def write():
        sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
else:
    rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())
rrmm = lambda n: [rrm() for _ in xrange(n)]

####

def solve(N, A):
    def kadane(B):
        ans = cur = 0
        for x in B:
            cur = max(cur + x, x)
            if cur>ans:ans=cur
        return ans
    
    S = sum(A[::2])
    B1 = []
    B2 = []
    for i in xrange(0, N-1, 2):
        B1.append(A[i+1] - A[i])
    for i in xrange(1, N-1, 2):
        B2.append(-A[i+1] +A[i])
    shift1 = kadane(B1)
    shift2 = kadane(B2)
    #print("!", B1, B2)
    #print(shift1, shift2)
    return max(S + shift1, S + shift2, S)

for T in xrange(rri()):
    N = rri()
    A = rrm()
    print solve(N, A)