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 case in range(t):
    n = int(input())
    a = list(map(int, input().split()))

    # prefix sum for how much you would gain if you switched this elements
    #su = 0
    #prefix_sum = [0]
    current_su = 0
    for i in range(n):
        #i2 = i + 1
        #su += (a[i2] - a[i])
        #prefix_sum.append(su)
        if (i % 2 == 0):
            current_su += a[i]
    #print (prefix_sum)
    
    #single_switches = [0]
    odd_su = 0
    even_su = 0
    pre_odd = [0]
    pre_even = [0]
    for i in range(n - 1):
        switch = (a[i + 1 - i % 2] - a[i + i % 2])
        if (i % 2 == 1):
            odd_su += switch
            pre_odd.append(odd_su)
        else:
            even_su += switch
            pre_even.append(even_su)
    
    smallest_odd = 0
    smallest_even = 0
    best = 0
    for i in range(len(pre_odd)):
        val = pre_odd[i]
        best = max(best, val - smallest_odd)
        smallest_odd = min(smallest_odd, val)
    
    for i in range(len(pre_even)):
        val = pre_even[i]
        best = max(best, val - smallest_even)
        smallest_even = min(smallest_even, val)
    
    #print (single_switches)
    # still have to account for whole subset switches starting at odd numbers
    #print (current_su)
    print (current_su + best)