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 solve(arr,n,ans):
    odd = 0
    even = 0
    dp = []
    even_indices = [(float('inf'),-1)]
    odd_indices = [(float('inf'),-1)]
    for i in range(n):
        if i%2 == 0:
            even += arr[i]
        else:
            odd += arr[i]

        dp.append(odd-even)
        if i%2 == 0:
            odd_indices.append(odd_indices[-1])
            if dp[-1] <= even_indices[-1][0]:
                even_indices.append((dp[-1],i))
            else:
                even_indices.append(even_indices[-1])
        else:
            even_indices.append(even_indices[-1])
            if dp[-1] <= odd_indices[-1][0]:
                odd_indices.append((dp[-1],i))
            else:
                odd_indices.append(odd_indices[-1])

    even_indices.pop(0)
    odd_indices.pop(0)
    max_val = 0
    l = -1
    r = -1
    for i in range(1,n):
        if i%2 != 0:
            diff = dp[i]-odd_indices[i-1][0]
            if diff > max_val:
                max_val = diff
                l = odd_indices[i-1][1]+1
                r = i

            if dp[i] > max_val:
                max_val = dp[i]
                l = 0
                r = i
        else:
            diff = dp[i]-even_indices[i-1][0]
            if diff > max_val:
                max_val = diff
                l = even_indices[i-1][1]+1
                r = i

    if l != -1:
        reverse(arr,l,r,n)

    total = find(arr,n)
    ans.append(total)

def reverse(arr,l,r,n):
    for i in range((r-l+1)//2):
        arr[l+i],arr[r-i] = arr[r-i],arr[l+i]


def find(arr,n):
    total = 0
    for i in range(n):
        if i%2 == 0:
            total += arr[i]

    #print(arr)
    return total

def brute(arr,n,ans):
    ans = 0
    li = -1
    ri = -1
    for l in range(n):
        for r in range(l,n):
            reverse(arr,l,r,n)
            total = find(arr,n)
            if total > ans:
                ans = total
                li = l
                ri = r
            reverse(arr,l,r,n)

    #print(li,ri,ans)
            

def main():
    t = int(input())
    ans = []
    for i in range(t):
        n = int(input())
        arr = list(map(int,input().split()))
        solve(arr,n,ans)

    for i in ans:
        print(i)


main()
