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
# cook your dish here
#code
import math
import collections
from sys import stdin,stdout,setrecursionlimit
from bisect import bisect_left as bsl
from bisect import bisect_right as bsr
import heapq as hq
setrecursionlimit(2**20)

t = 1
t = int(stdin.readline())

for _ in range(t):
    n = int(stdin.readline())
    
    a = list(map(int, stdin.readline().rstrip().split()))
    
    b = []
    ans = 0
    for i in range(0,n-1,2):
        b.append(a[i+1]-a[i])
        
    for i in range(n):
        if(i%2==0):
            ans += a[i]
    
    d = []
    for i in range(1,n-1,2):
        d.append(a[i]-a[i+1])
        
    #print(d)
    
    max_so_far = 0
    max_ending_here = 0
      
    for i in range(0, len(b)): 
        max_ending_here = max_ending_here + b[i] 
        if max_ending_here < 0: 
            max_ending_here = 0
        elif (max_so_far < max_ending_here): 
            max_so_far = max_ending_here
            
    max_so_far1 = 0
    max_ending_here1 = 0
      
    for i in range(0, len(d)): 
        max_ending_here1 = max_ending_here1 + d[i] 
        if max_ending_here1 < 0: 
            max_ending_here1 = 0
        elif (max_so_far1 < max_ending_here1): 
            max_so_far1 = max_ending_here1
            
            
    tmp = max(0,max_so_far,max_so_far1)
    print(ans+tmp)
    
    
    