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
__MULTITEST = True

def solve():
    n = int(input())
    a = list(map(int, input().split()))

    l = 0; r = -1;

    da = []
    db = []

    for i in range(1, n, 2):
        da.append(a[i] - a[i-1])
    
    glMax = 0
    la = 0; ra = -1;
    curMax = 0
    lc = 0; rc = -1;

    for i in range(len(da)):
        curMax += da[i];
        if curMax > 0:
            rc = i;
            if curMax > glMax:
                glMax = curMax
                la, ra = lc, rc;
        else:
            curMax = 0;
            lc = i + 1;

    db = []
    for i in range(2, n, 2):
        db.append(a[i-1] - a[i])

    glMaxB = 0
    lB = 0; rB = -1;
    curMax = 0
    lc = 0; rc = -1;

    for i in range(len(db)):
        curMax += db[i];
        if curMax > 0:
            rc = i;
            if curMax > glMaxB:
                glMaxB = curMax
                lB, rB = lc, rc;
        else:
            curMax = 0;
            lc = i + 1;
    
    if glMax < glMaxB:
        l = lB*2 + 1; r = rB*2 + 2;
    else:
        l = la*2; r = ra*2 + 1;

    a1 = a[:l]
    a2 = a[l:r+1]; a2.reverse()
    a3 = a[r+1:]
    #print(a1, a2, a3)
    a = a1 + a2 + a3
    #print(a)
    print(sum([a[i] for i in range(len(a)) if i % 2 == 0]))


# main
if __name__ == "__main__":
    t = (int(input()) if __MULTITEST else 1)
    for tt in range(t):
        solve()