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
for t in range(int(input())):
    n = int(input())
    *a, = map(int, input().split())

    dp = [[0]*3 for _ in range(n+1)]
    for i in range(n):
        # 反転を開始していない
        dp[i+1][0] = dp[i][0] + (a[i] if i % 2 == 0 else 0)

        # 反転開始したが、終了していない
        if i > 0:
            dp[i+1][1] = max(dp[i+1][1], max(dp[i-1][0], dp[i-1][1]) + (a[i] if (i-1) % 2 == 0 else a[i-1]))

        # 反転終了している
        dp[i+1][2] = max(dp[i][1], dp[i][2])+(a[i] if i % 2 == 0 else 0)

    print(max(dp[n]))