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
#!usr/bin/env python3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    t = I()
    for _ in range(t):
        n = I()
        a = LI()
        odd = [a[i] if i&1 else 0 for i in range(n)]
        even = [a[i]-odd[i] for i in range(n)]
        odd = [0]+list(accumulate(odd))
        even = [0]+list(accumulate(even))
        sub = [even[i]-odd[i] for i in range(n+1)]
        ans = max(even)
        l = 0
        k = even[1]-odd[1]
        for r in range(2,n,2):
            s = even[n]-even[r+1]+even[l+1]+odd[r+1]-odd[l+1]
            if ans < s:
                ans = s
            p = sub[r+1]
            if k < p:
                l = r
                k = p
        l = -1
        k = 0
        for r in range(1,n,2):
            s = even[n]-even[r+1]+even[l+1]+odd[r+1]-odd[l+1]
            if ans < s:
                ans = s
            p = sub[r+1]
            if k < p:
                l = r
                k = p
        print(ans)

    return

#Solve
if __name__ == "__main__":
    solve()
