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
import sys
ii = lambda: sys.stdin.readline().strip()
idata = lambda: [int(x) for x in ii().split()]

def ad(asa):
    ans = 0
    if bool(asa):
        ans = asa[0]
        summ = 0
        min_sum = 0
        for r in range(len(asa)):
            summ += asa[r]
            ans = max(ans, summ - min_sum)
            min_sum = min(min_sum, summ)
    return ans

def solve():
    n = int(ii())
    data = idata()
    if n == 1:
        print(data[0])
        return
    ans = 0
    for i in range(0, n, 2):
        ans += data[i]
    sp1 = []
    sp2 = []
    for i in range(0, n - 1, 2):
        sp1 += [data[i + 1] - data[i]]
        if i + 2 != n:
            sp2 += [data[i + 1] - data[i + 2]]
    print(max(ans, ans + ad(sp1), ans + ad(sp2)))
    return

for t in range(int(ii())):
    solve()
