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
#include <bits/stdc++.h>
using namespace std;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int main() {
  setIO();
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long arr[n], sum = 0;
    for (int i = (0); i < (n); ++i) {
      cin >> arr[i];
      if (i % 2 == 0) sum += arr[i];
    }
    long long MX = 0;
    multiset<long long> even, odd;
    even.insert(0);
    odd.insert(0);
    long long runningEven = 0, runningOdd = 0;
    for (int i = (0); i < (n); ++i) {
      if (i % 2 == 0) {
        runningEven += arr[i];
        MX = max(runningOdd - runningEven - *even.begin(), MX);
        even.insert(runningOdd - runningEven);
      } else {
        runningOdd += arr[i];
        MX = max(runningOdd - runningEven - *odd.begin(), MX);
        odd.insert(runningOdd - runningEven);
      }
    }
    cout << max(sum, sum + MX) << '\n';
  }
}
