#include <bits/stdc++.h>
using namespace std;
void _main(void) {
  int n;
  cin >> n;
  vector<long long> arr;
  do {
    arr.resize(n);
    for (auto& a : arr) cin >> a;
  } while (false);
  vector<long long> diff;
  for (int i = 1; i + 1 < n; i += 2) {
    diff.emplace_back(arr[i] - arr[i + 1]);
  }
  long long one = 0;
  if (diff.size()) {
    vector<long long> memo(diff.size());
    memo[0] = diff[0];
    one = memo[0];
    for (int i = 1; i < (int)diff.size(); ++i) {
      memo[i] = ((memo[i - 1] + diff[i]) > (diff[i]) ? (memo[i - 1] + diff[i])
                                                     : (diff[i]));
      one = ((one) > (memo[i]) ? (one) : (memo[i]));
    }
  }
  diff.clear();
  for (int i = 0; i + 1 < n; i += 2) {
    diff.emplace_back(arr[i + 1] - arr[i]);
  }
  long long two = 0;
  if (diff.size()) {
    vector<long long> memo(diff.size());
    memo[0] = diff[0];
    two = memo[0];
    for (int i = 1; i < (int)diff.size(); ++i) {
      memo[i] = ((memo[i - 1] + diff[i]) > (diff[i]) ? (memo[i - 1] + diff[i])
                                                     : (diff[i]));
      two = ((two) > (memo[i]) ? (two) : (memo[i]));
    }
  }
  long long acc = 0;
  for (int i = 0; i < n; i += 2) {
    acc += arr[i];
  }
  cout << ((acc) > (acc + ((one) > (two) ? (one) : (two)))
               ? (acc)
               : (acc + ((one) > (two) ? (one) : (two))))
       << '\n';
  return;
}
int main(void) {
  cin.sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  do {
    _main();
  } while (--t);
  return EXIT_SUCCESS;
}

### 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;
void _main(void) {
  int n;
  cin >> n;
  vector<long long> arr;
  do {
    arr.resize(n);
    for (auto& a : arr) cin >> a;
  } while (false);
  vector<long long> diff;
  for (int i = 1; i + 1 < n; i += 2) {
    diff.emplace_back(arr[i] - arr[i + 1]);
  }
  long long one = 0;
  if (diff.size()) {
    vector<long long> memo(diff.size());
    memo[0] = diff[0];
    one = memo[0];
    for (int i = 1; i < (int)diff.size(); ++i) {
      memo[i] = ((memo[i - 1] + diff[i]) > (diff[i]) ? (memo[i - 1] + diff[i])
                                                     : (diff[i]));
      one = ((one) > (memo[i]) ? (one) : (memo[i]));
    }
  }
  diff.clear();
  for (int i = 0; i + 1 < n; i += 2) {
    diff.emplace_back(arr[i + 1] - arr[i]);
  }
  long long two = 0;
  if (diff.size()) {
    vector<long long> memo(diff.size());
    memo[0] = diff[0];
    two = memo[0];
    for (int i = 1; i < (int)diff.size(); ++i) {
      memo[i] = ((memo[i - 1] + diff[i]) > (diff[i]) ? (memo[i - 1] + diff[i])
                                                     : (diff[i]));
      two = ((two) > (memo[i]) ? (two) : (memo[i]));
    }
  }
  long long acc = 0;
  for (int i = 0; i < n; i += 2) {
    acc += arr[i];
  }
  cout << ((acc) > (acc + ((one) > (two) ? (one) : (two)))
               ? (acc)
               : (acc + ((one) > (two) ? (one) : (two))))
       << '\n';
  return;
}
int main(void) {
  cin.sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  do {
    _main();
  } while (--t);
  return EXIT_SUCCESS;
}

```