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>
#pragma GCC optimize("Ofast")
using namespace std;
long long unsigned solve() {
  int n;
  cin >> n;
  vector<int> even, odd;
  even.reserve((n + 1) / 2);
  odd.reserve(n / 2);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (i & 1)
      odd.push_back(x);
    else
      even.push_back(x);
  }
  long long unsigned not_inverted = 0, inverting = 0, inverted = 0;
  for (int i = 0; i < odd.size(); i++) {
    inverted = max(inverted + even[i],
                   max(not_inverted + even[i], inverting + even[i]));
    inverting = max(inverting + odd[i], not_inverted + odd[i]);
    not_inverted += even[i];
  }
  long long unsigned ans = max(not_inverted, max(inverting, inverted));
  if (odd.size() != even.size()) {
    ans += even.back();
  }
  not_inverted = even[0], inverting = even[0], inverted = even[0];
  for (int i = 1; i < even.size(); i++) {
    inverted = max(inverted + even[i],
                   max(not_inverted + even[i], inverting + even[i]));
    inverting = max(inverting + odd[i - 1], not_inverted + odd[i - 1]);
    not_inverted += even[i];
  }
  return max(ans, max(not_inverted, max(inverting, inverted)));
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  for (int test_case = 0; test_case < t; test_case++) {
    cout << solve() << "\n";
  }
  return 0;
}
