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
n = int(input())
while n > 0:
    arr = []
    t = int(input())
    arr = [int(tp) for tp in input().split()]
    ans = 0
    for i in range(t):
        if not i % 2:
            ans += arr[i]
    '''dp = [[0 for i in range(t + 1)] for j in range(t)]
    for i in range(2,t + 1,2):
        for j in range(t - i + 1):
            if i == 2:
                tp = arr[j + 1] - arr[j] if j % 2 == 0 else arr[j] - arr[j + 1]
                dp[j][i] = max(0 , tp)
            else:
                tp = 0 
                for k in range(j , j + i):
                    if k % 2 == 0 and (k + 1) < j + i:
                        tp += arr[k + 1] - arr[k]
                dp[j][i] = max(dp[j][i - 2] , dp[j + 1][i - 2] , dp[j + 2][i - 2] , tp) 
    if t % 2:
        ans += max(dp[0][t - 1] , dp[1][t - 1]) 
    else:
        ans += dp[0][t]'''
    dp , maxn = 0 , 0
    for i in range(0 , t , 2):
        if i + 1 < t:
            dp = max(dp + arr[i + 1] - arr[i] , 0)
            maxn = max(dp , maxn)
    dp = 0
    for i in range(1 , t , 2):
        if i + 1 < t:
            dp = max(dp + arr[i] - arr[i + 1] , 0)
            maxn = max(dp , maxn)
    print(ans + maxn)
    n -= 1