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 sys import stdin

t = int(stdin.readline())

for _ in range(t):
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().rstrip().split(" ")))
    prefixArr = [0]
    out = 0
    for i in range(1, n):
        if i % 2 != 0:
            diff = a[i] - a[i - 1]
        else:
            diff = a[i - 1] - a[i]
        
        if diff > diff + prefixArr[max(0, i - 2)]:
                prefixArr.append(diff)
        else:
            prefixArr.append(diff + prefixArr[max(0, i - 2)])

    for i in range(n):
        if i % 2 == 0:
            out += a[i]
    # print('=====')
    # print(out)
    # print(prefixArr)
    # print('=====')

    print(out + max(prefixArr))    
