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
t = int(input())
for u in range(t):
    n = int(input())
    a = list(map(int, input().split(' ')))
    b = [a[i] - a[i - 1] for i in range(1, n, 2)]
    b.append(0)
    m1 = [b[0]]
    for i in range(1, b.__len__()):
        if m1[i - 1] >= 0:
            m1.append(m1[i - 1] + b[i])
        else:
            m1.append(b[i])
    ans = sum([a[i] for i in range(n) if (i - 1) & 1]) + max(max(m1), 0)
    c = [a[i-1] - a[i] for i in range(2, n, 2)]
    c.append(0)
    m2 = [c[0]]
    for i in range(1, c.__len__()):
        if m2[i - 1] >= 0:
            m2.append(m2[i - 1] + c[i])
        else:
            m2.append(c[i])
    ans = max(ans, sum([a[i] for i in range(n) if (i - 1) & 1]) + max(max(m2), 0))

    print(ans)
