#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n + 1);
  vector<long long> even_sum(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    even_sum[i] = even_sum[i - 1];
    if (i % 2) even_sum[i] += a[i];
  }
  if (n == 1) {
    cout << a[1] << '\n';
    return;
  }
  vector<int> b(1, 0);
  for (int i = 2; i <= n; i += 2) b.push_back(a[i] - a[i - 1]);
  long long ans = b[1], sum = 0;
  for (size_t r = 1; r < b.size(); ++r) {
    sum += b[r];
    ans = max(ans, sum);
    sum = max(sum, 0ll);
  }
  long long best = even_sum[n] + max(ans, 0ll);
  if (n > 2) {
    vector<int> c(1, 0);
    for (size_t i = 3; i < a.size(); i += 2) c.push_back(a[i - 1] - a[i]);
    ans = c[1], sum = 0;
    for (size_t r = 1; r < c.size(); ++r) {
      sum += c[r];
      ans = max(ans, sum);
      sum = max(sum, 0ll);
    }
    best = max(best, even_sum[n] + ans);
  }
  cout << best << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP to 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;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n + 1);
  vector<long long> even_sum(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    even_sum[i] = even_sum[i - 1];
    if (i % 2) even_sum[i] += a[i];
  }
  if (n == 1) {
    cout << a[1] << '\n';
    return;
  }
  vector<int> b(1, 0);
  for (int i = 2; i <= n; i += 2) b.push_back(a[i] - a[i - 1]);
  long long ans = b[1], sum = 0;
  for (size_t r = 1; r < b.size(); ++r) {
    sum += b[r];
    ans = max(ans, sum);
    sum = max(sum, 0ll);
  }
  long long best = even_sum[n] + max(ans, 0ll);
  if (n > 2) {
    vector<int> c(1, 0);
    for (size_t i = 3; i < a.size(); i += 2) c.push_back(a[i - 1] - a[i]);
    ans = c[1], sum = 0;
    for (size_t r = 1; r < c.size(); ++r) {
      sum += c[r];
      ans = max(ans, sum);
      sum = max(sum, 0ll);
    }
    best = max(best, even_sum[n] + ans);
  }
  cout << best << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```