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

# import math
from collections import deque

# import heapq
# from math import inf
# from math import gcd

# print(help(deque))
# 26
pprint = lambda s: print(' '.join(map(str, s)))
input = lambda: sys.stdin.readline().strip()
ipnut = input

# n,x = map(int,input().split())
for i in range(int(input())):
    n = int(input())
    a = list(map(int,input().split()))
    d = sum(a[::2])
    ans = d
    for i in range(n):
        if i%2==0:
            a[i] = -a[i]
    l,r = 0,1
    curr_sum = 0
    while r<n:
        curr_sum+=a[r]+a[r-1]
        r+=2
        if curr_sum<0:
            while curr_sum<0 or l<r-3:
                curr_sum-=(a[l]+a[l+1])
                l+=2
        ans = max(ans,d+curr_sum)
    l, r = 1, 2
    curr_sum = 0
    while r < n:
        curr_sum += a[r] + a[r - 1]
        r += 2
        if curr_sum < 0:
            while curr_sum < 0 or l < r - 3:
                curr_sum -= (a[l] + a[l + 1])
                l += 2
        ans = max(ans, d + curr_sum)
    print(ans)