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
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')




def solve(n, a):
    base = 0
    for i in range(0, n, 2):
        base += a[i]

    ans = base

    # even indexes
    l = 0
    best = 0
    s = 0
    for r in range(n // 2):
        s += (a[2*r+1] - a[2*r])
        while s < 0:
            s -= (a[2*l+1] - a[2*l])
            l += 1
        if s > best:
            best = s
            ans = max(ans, base + best)

    # odd indexes
    l = 0
    best = 0
    s = 0
    for r in range((n-1) // 2):
        s += (a[2 * r + 1] - a[2 * r + 2])
        while s < 0:
            s -= (a[2 * l + 1] - a[2 * l + 2])
            l += 1
        if s > best:
            best = s
            ans = max(ans, base + best)

    return ans


def main():
    for _ in range(ri()):
        n = ri()
        a = ria()
        wi(solve(n, a))


if __name__ == '__main__':
    main()
