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

from collections import defaultdict


for i in range(int(input())):
    n = int(input())
    arr = [int(i) for i in input().split()]
    eo = [[0,0]]
    even = 0
    odd = 0
    

    mn_o = [10**3]
    mn_e = [10**3]

    
    for i in range(len(arr)):
        if i%2 == 0:
            even += arr[i]
        else:
            odd += arr[i]
        eo.append([even, odd])
    

    for i in range(len(eo)):
        if i%2 == 0:
            mn_e.append(min(mn_e[-1], eo[i][1]-eo[i][0]))
            mn_o.append(mn_o[-1])
        else:
            mn_o.append(min(mn_o[-1],eo[i][1]-eo[i][0]))
            mn_e.append(mn_e[-1])

    ans = 0
    for i in range(1, len(eo), 2):
        ans = max(ans, eo[i][1]-eo[i][0] - mn_o[i])

    for i in range(0, len(eo), 2):
        ans = max(ans, eo[i][1]-eo[i][0] - mn_e[i])
    print(ans+eo[-1][0])