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 _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    
    if n <= 2:
        print(max(a))
    else:
        s = sum(a[::2])
        diff1 = []
        for i in range(1, n, 2):
            diff1.append(a[i]-a[i-1])
            
        diff2 = []
        for i in range(2, n, 2):
            diff2.append(a[i-1]-a[i])
        
        msf1 = max(diff1[0], 0)
        ans1 = msf1
        for i in range(1, len(diff1)):
            msf1 = diff1[i] + max(0, msf1)
            ans1 = max(msf1, ans1)
            
        msf2 = max(diff2[0], 0)
        ans2 = msf2
        for i in range(1, len(diff2)):
            msf2 = diff2[i] + max(0, msf2)
            ans2 = max(msf2, ans2)
            
        print(s+max(ans1, ans2))