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 typing import *

import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq


def input():
    return sys.stdin.readline()[:-1]


# sys.setrecursionlimit(1000000)

# _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

# """
# sys.stdin = io.StringIO(_INPUT)

INF = 10**10

def calc(T):
    N = len(T)

    # 後ろからの累積max
    Tm = [-INF] * N
    Tm[N-1] = T[N-1]
    for i in reversed(range(N-1)):
        Tm[i] = max(Tm[i+1], T[i])
    return max(Tm[i]-T[i] for i in range(N))

def solve(N, A):
    S0 = [0]
    S1 = [0]
    for i in range(N):
        if i % 2 == 0:
            S0.append(S0[-1] + A[i])
            S1.append(S1[-1])
        else:
            S0.append(S0[-1])
            S1.append(S1[-1] + A[i])
    T = [S1[i] - S0[i] for i in range(N+1)]

    T0 = T[0::2]
    T1 = T[1::2]
    v = max(calc(T0), calc(T1))

    return S0[N] + v

T0 = int(input())
for _ in range(T0):
    N = int(input())
    A = list(map(int, input().split()))
    print(solve(N, A))