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

def maxSub(a) :
  mxTot = a[0]
  mxHere = a[0]
  for i in range(1,len(a)) :
    mxHere = max(a[i], mxHere + a[i])
    mxTot = max(mxTot, mxHere)
  return mxTot

def solve(N, arr) :

  cur = 0
  changes1 = []
  for n in range(0,N,2) :
    if n+1 < N : changes1.append(arr[n+1]-arr[n])
    cur += arr[n]

  changes2 = []
  for n in range(1,N,2) :
    if n+1 < N : changes2.append(arr[n]-arr[n+1])

  mx1, mx2 = 0, 0
  if changes1 : mx1 = maxSub(changes1)
  if changes2 : mx2 = maxSub(changes2)

  return max(mx1,mx2,0) + cur


T = input()
for t in range(T) :
  N = input()
  arr = map(int,raw_input().split())

  ans = solve(N,arr)
  if N%2 == 1 : ans = max(ans, solve(N,arr[::-1]))

  print ans


