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 atexit, io, sys 
  
# A stream implementation using an in-memory bytes  
# buffer. It inherits BufferedIOBase. 
buffer = io.BytesIO() 
sys.stdout = buffer
  
# print via here 
@atexit.register 
def write(): 
    sys.__stdout__.write(buffer.getvalue())

for _ in range(input()):
    n=input()
    
    a=map(int,raw_input().split())
    if n==1:
        print a[0];continue
    
    dp=[[-float('inf'),-float('inf'),0]for j in range (n)]
    dp[0][2]=a[0]
    dp[1][2]=a[0]
    for j in range(2,n):
        if j%2:
            dp[j][2]=dp[j-1][2]
        else:
            dp[j][2]=a[j]+dp[j-1][2]
         
    
    dp[0][0]=a[0]
    dp[0][1]=a[0]
    dp[1][0]=a[0]
    dp[1][1]=a[1]
    
    for j in range(2,n):
        if (j%2):
            dp[j][0]=max(dp[j-1][0],dp[j-1][1])
            dp[j][1]=a[j]+max(dp[j-2][2],dp[j-2][1])
        else:
            dp[j][0]=max(dp[j-1][0],dp[j-1][1])+a[j]
            dp[j][1]=a[j-1]+max(dp[j-2][2],dp[j-2][1])
    print max([dp[n-1][0],dp[n-1][1],dp[n-1][2]])
   