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
for _ in range(int(input())):
    n=int(input())
    a=list(map(int,input().split()))
    dp0=[[0,0,0] for i in range((n-1)//2+1)]
    for i in range((n-1)//2+1):
        if i==0:
            dp0[i][0]=a[2*i]
            continue
        dp0[i][0]=dp0[i-1][0]+a[2*i]
        dp0[i][1]=max(dp0[i-1][0],dp0[i-1][1])+a[2*i-1]
        dp0[i][2]=max(dp0[i-1][1],dp0[i-1][2])+a[2*i]
    dp1=[[0,0,0] for i in range((n-1)//2+1)]
    for i in range((n-1)//2+1):
        if i==0:
            dp1[i][0]=a[2*i]
            if 2*i+1<n:dp1[i][1]=a[2*i+1]
            continue
        dp1[i][0]=dp1[i-1][0]+a[2*i]
        if 2*i+1<n:dp1[i][1]=max(dp1[i-1][0],dp1[i-1][1])+a[2*i+1]
        if 2*i<n:dp1[i][2]=max(dp1[i-1][1],dp1[i-1][2])+a[2*i]
    #print(dp0)
    #print(dp1
    print(max(dp0[(n-1)//2]+dp1[(n-1)//2]))