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, setrecursionlimit, stdout
#setrecursionlimit(1000000)
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
from heapq import heapify, heappop, heappush, heappushpop, heapreplace
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline().rstrip()
def lsi(): return list(si())
#mod=1000000007
res=['NET', 'DA']
############# CODE STARTS HERE #############


test_case=ii()
while test_case:
    test_case-=1



    n=ii()
    a=li()
    s=sum([a[i] for i in range(0, n, 2)])
    s1=s2=mx1=mx2=0
    for i in range(1, n, 2):
        x=a[i]-a[i-1]
        if s1+x>0:
            s1+=x
            mx1=max(mx1, s1)
        else:
            s1=0

    for i in range(2, n, 2):
        y=a[i-1]-a[i]
        if s2+y>0:
            s2+=y
            mx2=max(mx2, s2)
        else:
            s2=0

    print(s+max(mx1, mx2))