#include <bits/stdc++.h>
using namespace std;
const long long int N = 200005;
long long int a[N], p[N], pe[N];
set<long long int> o, e;
void solve() {
  o.clear();
  e.clear();
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  p[0] = a[0];
  pe[0] = a[0];
  o.insert(0);
  e.insert(a[0]);
  long long int mx = 0;
  for (long long int i = 1; i < n; i++) {
    p[i] = p[i - 1] + a[i];
    if (i % 2)
      pe[i] = pe[i - 1];
    else
      pe[i] = pe[i - 1] + a[i];
    long long int x = 2 * pe[i] - p[i];
    if (i % 2) {
      long long int y = *(o.rbegin());
      y -= x;
      mx = max(mx, y);
    } else {
      long long int y = *(e.rbegin());
      y -= x;
      mx = max(mx, y);
    }
    if (i % 2)
      o.insert(x);
    else
      e.insert(x);
  }
  cout << pe[n - 1] + mx << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cout.precision(20);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    if (0) cout << "Case #" << i << ": ";
    solve();
  }
}

### 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;
const long long int N = 200005;
long long int a[N], p[N], pe[N];
set<long long int> o, e;
void solve() {
  o.clear();
  e.clear();
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  p[0] = a[0];
  pe[0] = a[0];
  o.insert(0);
  e.insert(a[0]);
  long long int mx = 0;
  for (long long int i = 1; i < n; i++) {
    p[i] = p[i - 1] + a[i];
    if (i % 2)
      pe[i] = pe[i - 1];
    else
      pe[i] = pe[i - 1] + a[i];
    long long int x = 2 * pe[i] - p[i];
    if (i % 2) {
      long long int y = *(o.rbegin());
      y -= x;
      mx = max(mx, y);
    } else {
      long long int y = *(e.rbegin());
      y -= x;
      mx = max(mx, y);
    }
    if (i % 2)
      o.insert(x);
    else
      e.insert(x);
  }
  cout << pe[n - 1] + mx << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cout.precision(20);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    if (0) cout << "Case #" << i << ": ";
    solve();
  }
}

```