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
# input = open('file.txt').readline

def kadane( pref ):
  if pref:

    mxg = mxl = pref[0]
    pref.pop(0)
    for p in pref:

      mxl = max( p , p + mxl )

      if mxl > mxg:
        mxg = mxl
    
    
    if mxg < 0:
      mxg = 0

    return mxg
  
  return 0

for _ in range( int(input()) ):

  n = int(input())
  arr = list( map( int , input().strip().split(' ') ) )

  evesum = 0
  pref = []
  suf = []
  for i in range(0 , n , 2):
    evesum += arr[i]
    
    if i+1 < n:
      pref.append( arr[i+1] - arr[i] )
    if i > 1:
      suf.append( arr[i-1] - arr[i] )
  
  # print('eves', evesum , 'pref' , pref , suf)

  extra = max( kadane( pref ) , kadane( suf ) )

  print( evesum + extra )
  

