#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
template <typename T>
ostream &operator+(ostream &out, const vector<T> &vec) {
  for (const auto &x : vec) {
    out << x << " ";
  }
  out << "\n";
  return out;
}
template <typename T>
ostream &operator*(ostream &out, const vector<T> &vec) {
  for (const auto &x : vec) {
    out + x;
  }
  return out;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &vec) {
  for (auto &x : vec) {
    in >> x;
  }
  return in;
}
ll kadane(const vector<int> &seq) {
  ll currans = 0;
  ll ans = 0;
  for (const auto &val : seq) {
    currans += val;
    if (currans < 0) {
      currans = 0;
    }
    ans = max(currans, ans);
  }
  return ans;
}
void solve() {
  int n;
  cin >> n;
  vector<int> seq(n);
  cin >> seq;
  ll ans = 0;
  for (int i = 0; i < n; i += 2) {
    ans += seq[i];
  }
  vector<int> kadane1;
  vector<int> kadane2;
  for (int i = 0; i + 1 < n; i += 2) {
    kadane1.push_back(seq[i + 1] - seq[i]);
  }
  for (int i = 1; i + 1 < n; i += 2) {
    kadane2.push_back(seq[i] - seq[i + 1]);
  }
  ans = max({ans, ans + kadane(kadane1), ans + kadane(kadane2)});
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Create a solution in cpp for the following problem:

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;
using ll = long long int;
template <typename T>
ostream &operator+(ostream &out, const vector<T> &vec) {
  for (const auto &x : vec) {
    out << x << " ";
  }
  out << "\n";
  return out;
}
template <typename T>
ostream &operator*(ostream &out, const vector<T> &vec) {
  for (const auto &x : vec) {
    out + x;
  }
  return out;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &vec) {
  for (auto &x : vec) {
    in >> x;
  }
  return in;
}
ll kadane(const vector<int> &seq) {
  ll currans = 0;
  ll ans = 0;
  for (const auto &val : seq) {
    currans += val;
    if (currans < 0) {
      currans = 0;
    }
    ans = max(currans, ans);
  }
  return ans;
}
void solve() {
  int n;
  cin >> n;
  vector<int> seq(n);
  cin >> seq;
  ll ans = 0;
  for (int i = 0; i < n; i += 2) {
    ans += seq[i];
  }
  vector<int> kadane1;
  vector<int> kadane2;
  for (int i = 0; i + 1 < n; i += 2) {
    kadane1.push_back(seq[i + 1] - seq[i]);
  }
  for (int i = 1; i + 1 < n; i += 2) {
    kadane2.push_back(seq[i] - seq[i + 1]);
  }
  ans = max({ans, ans + kadane(kadane1), ans + kadane(kadane2)});
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```