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
import sys
import bisect as b
import math
from collections import defaultdict as dd
input=sys.stdin.readline
#sys.setrecursionlimit(10**7)
mo=10**9+7
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
def pref(a,n,f):             
    pre=[0]*n
    if(f==0):         ##from beginning
        pre[0]=a[0]
        for i in range(1,n):
            pre[i]=a[i]+pre[i-1]
    else:              ##from end
        pre[-1]=a[-1]
        for i in range(n-2,-1,-1):
            pre[i]=pre[i+1]+a[i]
    return pre
maxint=10**24 
def kadane(a,size): 
       
    max_so_far = -maxint - 1
    max_ending_here = 0
       
    for i in range(0, size): 
        max_ending_here = max_ending_here + a[i] 
        if (max_so_far < max_ending_here): 
            max_so_far = max_ending_here 
  
        if max_ending_here < 0: 
            max_ending_here = 0   
    return max_so_far 
for _ in range(inin()):
    n=inin()
    a=ain()
    ans=sum(a[0:n:2])
    l3=[]
    l4=[]
    for i in range(0,n-1,2):
        l3+=[a[i+1]-a[i]]
    for i in range(1,n-1,2):
        l4+=[a[i]-a[i+1]]
    print(max(kadane(l3,len(l3)),kadane(l4,len(l4)),0)+ans)
            
    
    

        
        
    
                
                
                
                

        
            
            
                
            
                
    
 
        
    
    
    
    




    


   
        
    
    
    
    




    


