#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18 + 3;
const int N = 1e5 + 1;
const int MOD = 998244353;
const double EPS = 1e-9;
const double PI = acos(-1);
long long max_segment_sum(vector<int> a) {
  long long ans = a[0], sum = 0;
  for (int i = 0; i < a.size(); i++) {
    sum += a[i];
    ans = max(ans, sum);
    sum = max(sum, 0ll);
  }
  return ans;
}
void solve() {
  int n;
  cin >> n;
  int a[n];
  auto sum = 0ll;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i] * !(i & 1);
  }
  vector<int> kek;
  for (int i = 0; i < n - 1; i += 2) {
    kek.push_back(a[i + 1] - a[i]);
  }
  auto ans = 0ll;
  if (!kek.empty()) ans = max(ans, max_segment_sum(kek));
  kek.clear();
  for (int i = (n & 1 ? n - 1 : n - 2); i > 1; i -= 2) {
    kek.push_back(a[i - 1] - a[i]);
  }
  if (!kek.empty()) ans = max(ans, max_segment_sum(kek));
  cout << sum + ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18 + 3;
const int N = 1e5 + 1;
const int MOD = 998244353;
const double EPS = 1e-9;
const double PI = acos(-1);
long long max_segment_sum(vector<int> a) {
  long long ans = a[0], sum = 0;
  for (int i = 0; i < a.size(); i++) {
    sum += a[i];
    ans = max(ans, sum);
    sum = max(sum, 0ll);
  }
  return ans;
}
void solve() {
  int n;
  cin >> n;
  int a[n];
  auto sum = 0ll;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i] * !(i & 1);
  }
  vector<int> kek;
  for (int i = 0; i < n - 1; i += 2) {
    kek.push_back(a[i + 1] - a[i]);
  }
  auto ans = 0ll;
  if (!kek.empty()) ans = max(ans, max_segment_sum(kek));
  kek.clear();
  for (int i = (n & 1 ? n - 1 : n - 2); i > 1; i -= 2) {
    kek.push_back(a[i - 1] - a[i]);
  }
  if (!kek.empty()) ans = max(ans, max_segment_sum(kek));
  cout << sum + ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```