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 collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math

# sys.stdin = open('input')

def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x

mod = 1000000007

def my_main():
    T = inp()
    for i in range(T):
        n = inp()
        da = inp(True)
        ans = 0
        dp = []
        dp2 = []
        for i in range(n):
            if i%2==0:
                ans += da[i]
                if i>=2:
                    dp2.append(da[i-1]-da[i])
            else:
                dp.append(da[i] - da[i-1])

        mi = 0
        sm = 0
        tans = 0
        for i in dp:
            sm += i
            mi = min(sm, mi)
            tans = max(tans, sm-mi)
        r_ans = tans + ans
        mi = 0
        sm = 0
        tans = 0
        for i in dp2:
            sm += i
            mi = min(sm, mi)
            tans = max(tans, sm-mi)
        r_ans = max(r_ans, tans + ans)
        print r_ans








my_main()
