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 maxEvenLenSum(arr, n): 
    if (n < 2): 
        return 0
    dp = [0 for i in range(n)] 
    dp[n - 1] = 0
    dp[n - 2] = arr[n - 2] + arr[n - 1] 
  
    for i in range(n - 3, -1, -1): 
        dp[i] = arr[i] + arr[i + 1] 
        if (dp[i + 2] > 0): 
            dp[i] += dp[i + 2] 
    maxSum = max(dp) 
    return maxSum 
t=int(input())
for _ in range(t):
    n=int(input())
    A=[int(x) for x in input().split()]
    B=[]
    for i in range(n):
        if(i%2==0):
            B.append(A[i])
    for i in range(n):
        if(i%2==0):
            A[i]*=-1
    max_so_far=maxEvenLenSum(A,n)
    print(sum(B)+max_so_far)