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
cases = int(input())

for i in range(cases):
    n = int(input())
    l2 = input().split(' ')
    arr = []
    sum1 = 0    #to store original even indexes sum
    for a in l2:
        arr.append(int(a))

    #calculate pairwise difference
    diff1 = []
    for j in range(0,len(arr),2):
        if j+1 >= len(arr):
            break
        ele1 = arr[j]
        ele2 = arr[j+1]
        diff1.append(ele2-ele1)
    diff2 = []
    for k in range(1,len(arr),2):
        if k+1>=len(arr):
            break
        ele1 = arr[k]
        ele2 = arr[k+1]
        diff2.append(ele1-ele2)
    #calculate the original even indexes sum
    for p in range(len(arr)):
        if p%2 == 0:
            sum1 += arr[p]
    s1 = s2 = -1
    for a in range(len(diff1)):
        start = diff1[a]
        if start > 0:
            s1 = a
            break
    for b in range(len(diff2)):
        start = diff2[b]
        if start > 0:
            s2 = b
            break
    e1 = e2 = -1
    for c in range(len(diff1)-1,-1,-1):
        end = diff1[c]
        if end > 0:
            e1 = c
            break
    for d in range(len(diff2)-1,-1,-1):
        end = diff2[d]
        if end > 0:
            e2 = d
            break
    #if s1 exists then e1 must also exists, both as non negative indexes
    #if s2 exists then e2 must also exists, both as non negative indexes
    #s1 cannot be larger than e1
    #s2 cannot be larger than e2
    if s1 == e1 == s2 == e2 == -1:
        print (sum1)
        continue

    sum2 = sum3 = 0
    t1 = t2 = 0
    if e1 >= s1 and e1 != -1:
        for w in range(s1,e1+1):
            num = diff1[w]
            sum2 += num
            if sum2 > t1:
                t1 = sum2
            if sum2 <0:
                sum2 = 0
    if e2 >= s2 and e2 != -1:
        for x in range(s2,e2+1):
            num = diff2[x]
            sum3 += num
            if sum3>t2:
                t2 = sum3
            if sum3 < 0:
                sum3 = 0
    m = max(t1,t2)
    print (sum1 + m)

    #expected 18 found 17