#include <bits/stdc++.h>
using namespace std;
const long long MN = 2e5 + 7;
long long maxSubArraySum(vector<long long> a) {
  if (a.size() == 0) {
    return 0;
  }
  long long size = a.size();
  long long max_so_far = a[0];
  long long curr_max = a[0];
  for (long long i = 1; i < size; i++) {
    curr_max = max(a[i], curr_max + a[i]);
    max_so_far = max(max_so_far, curr_max);
  }
  return max_so_far;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long ans = 0;
    long long a[n + 1];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) {
        ans += a[i];
      }
    }
    vector<long long> ev, od;
    for (long long i = 0; i < n && i + 1 < n; i += 2) {
      ev.push_back(a[i + 1] - a[i]);
    }
    for (long long i = 1; i < n && i + 1 < n; i += 2) {
      od.push_back(a[i] - a[i + 1]);
    }
    long long k = max(maxSubArraySum(ev), maxSubArraySum(od));
    if (k > 0) {
      ans += k;
    }
    cout << ans << endl;
  }
}

### Prompt

Please provide a Cpp coded solution 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 MN = 2e5 + 7;
long long maxSubArraySum(vector<long long> a) {
  if (a.size() == 0) {
    return 0;
  }
  long long size = a.size();
  long long max_so_far = a[0];
  long long curr_max = a[0];
  for (long long i = 1; i < size; i++) {
    curr_max = max(a[i], curr_max + a[i]);
    max_so_far = max(max_so_far, curr_max);
  }
  return max_so_far;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long ans = 0;
    long long a[n + 1];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) {
        ans += a[i];
      }
    }
    vector<long long> ev, od;
    for (long long i = 0; i < n && i + 1 < n; i += 2) {
      ev.push_back(a[i + 1] - a[i]);
    }
    for (long long i = 1; i < n && i + 1 < n; i += 2) {
      od.push_back(a[i] - a[i + 1]);
    }
    long long k = max(maxSubArraySum(ev), maxSubArraySum(od));
    if (k > 0) {
      ans += k;
    }
    cout << ans << endl;
  }
}

```