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
input=stdin.readline
import bisect
import math
#i = bisect.bisect_left(a, k)
#list=input().split(maxsplit=1)
for xoxo in range(1):
    #a=[]
    for _ in range (int(input())):
        n=int(input())
        a=list(map(int, input().split()))
        if n==1:
            print(a[0])
            continue
        ans=[0,0]
        p,x=0,-1
        s=0
        for i in range(0,n,2):
            s+=a[i]
            if i+1 >= n:
                break
            p+=(a[i+1]-a[i])
            if x<p:
                x=p
            if p<0:
                p=0
        ans[0]=x
        p,x=0,-1
        for i in range(1,n,2):
            if i+1 >= n:
                break
            p+=(a[i]-a[i+1])
            if x<p:
                x=p
            if p<0:
                p=0
        ans[1]=x
        print(s+max(0,max(ans)))
            
    #ans=0  