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
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))

t,=I()

def ma(a,size): 
      
    max_so_far =a[0] 
    curr_max = a[0] 
      
    for i in range(1,size): 
        curr_max = max(a[i], curr_max + a[i]) 
        max_so_far = max(max_so_far,curr_max) 
          
    return max_so_far 

for _ in range(t):
	n,=I()
	l=I()
	pre=[0]*n
	suf=[0]*n
	an=0
	if n<=2:
		print(max(l))
		continue
	d=[]
	for i in range(n):
		if i>0:
			d.append(l[i-1]-l[i])
		if i%2==0:
			pre[i]=pre[max(0,i-2)]+l[i]
		if (n-i-1)%2:
			suf[n-i-1]=suf[min(n-1,n-i+1)]+l[n-i-1]
	#print(pre)
	#print(suf)
	x=[d[i] for i in range(n-1) if i%2 ]
	y=[-d[i] for i in range(n-1) if i%2==0]
	r=ma(x,len(x))
	p=ma(y,len(y))
	#print(x,r,y,p)
	r=max(r,p)
	if r<0:
		r=0
	an=max(max(pre[n-2],pre[n-1])+r,suf[1])
	for i in range(0,n-3,2):
		an=max(an,pre[i]+suf[i+3])
	print(an)
