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
"""
Code of Ayush Tiwari
Codeforces: servermonk
Codechef: ayush572000

"""
import sys
input = sys.stdin.buffer.readline

def solution():
    for _ in range(int(input())):
        
        # This is the main code
        n=int(input())
        l=list(map(int,input().split()))
        ecnt=0
        for i in range(0,n,2):
            ecnt+=l[i]
        pre1=[0]*(n)
        pre2=[0]*(n)
        for i in range(1,n,2):
            pre1[i]=l[i]-l[i-1]
            if i<n-1:
                pre2[i]=l[i]-l[i+1]
        # print(pre1)
        # print(pre2)
        pre1x=pre2x=pre1y=pre2y=0
        for i in range(n):
            pre1x+=pre1[i]
            pre2x+=pre2[i]
            if pre1x<0:
                pre1x=0
            if pre2x<0:
                pre2x=0
            if pre1x>pre1y:
                pre1y=pre1x
            if pre2x>pre2y:
                pre2y=pre2x
        print(ecnt+max(0,pre1y,pre2y))

solution()