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
from sys import stdin
input = stdin.readline


def answer():

    # dp[i][0] - > taking even
    # dp[i][1] - > extending odd from even
    # dp[i][2] - > extending odd from odd
    # dp[i][3] - > stopped extending odd

    dp = [[0 , 0 , 0 , 0] for i in range(n + 1)]
    for i in range(1 , n + 1):
        
        dp[i][0] = dp[i - 1][0]
        dp[i][1] = dp[i - 1][1]
        dp[i][2] = dp[i - 1][2]
        dp[i][3] = dp[i - 1][3]
        
        if(i & 1):
            dp[i][0] += a[i - 1]
            dp[i][3] += a[i - 1]
            dp[i][1] = max(dp[i][1] , dp[i - 1][0])

            dp[i][3] = max(dp[i][3] , dp[i][2] , dp[i][0])

        else:

            dp[i][1] += a[i - 1]
            dp[i][2] = max(dp[i][2] , dp[i - 1][0]) + a[i - 1]

            dp[i][3] = max(dp[i][3] , dp[i][1])
        
    return dp[n][3]
   

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

    n = int(input())

    a = list(map(int,input().split()))

    
    print( answer() )
