#include <bits/stdc++.h>
using namespace std;
long long kadane(vector<long long> temp) {
  long long res = 0, cur = 0;
  for (int i = 0; i < temp.size(); i++) {
    cur = max(cur + temp[i], temp[i]);
    res = max(cur, res);
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (long long &i : v) cin >> i;
  if (n == 1) {
    cout << v[0] << endl;
    return;
  }
  long long sum = 0;
  for (int i = 0; i < n; i += 2) sum += v[i];
  vector<long long> temp;
  for (int i = 0; i + 1 < n; i += 2) {
    if (i + 1 < n)
      temp.push_back(v[i + 1] - v[i]);
    else
      break;
  }
  long long summ = max(sum, sum + kadane(temp));
  temp.clear();
  for (int i = 1; i + 1 < n; i += 2) {
    if (i + 1 < n)
      temp.push_back(v[i] - v[i + 1]);
    else
      break;
  }
  sum = max(sum, sum + kadane(temp));
  cout << max(sum, summ) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your challenge is to write a CPP solution 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;
long long kadane(vector<long long> temp) {
  long long res = 0, cur = 0;
  for (int i = 0; i < temp.size(); i++) {
    cur = max(cur + temp[i], temp[i]);
    res = max(cur, res);
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (long long &i : v) cin >> i;
  if (n == 1) {
    cout << v[0] << endl;
    return;
  }
  long long sum = 0;
  for (int i = 0; i < n; i += 2) sum += v[i];
  vector<long long> temp;
  for (int i = 0; i + 1 < n; i += 2) {
    if (i + 1 < n)
      temp.push_back(v[i + 1] - v[i]);
    else
      break;
  }
  long long summ = max(sum, sum + kadane(temp));
  temp.clear();
  for (int i = 1; i + 1 < n; i += 2) {
    if (i + 1 < n)
      temp.push_back(v[i] - v[i + 1]);
    else
      break;
  }
  sum = max(sum, sum + kadane(temp));
  cout << max(sum, summ) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

```