#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
template <typename T>
void printarr(T *A, int n) {
  for (int i = 0; i < n; i++) {
    cout << A[i] << " ";
  }
  cout << "\n";
}
template <typename T>
void arrinit(T *A, int n, T data) {
  for (int i = 0; i < n; i++) {
    A[i] = data;
  }
}
template <typename T>
void takeinput(T *A, int n) {
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
}
template <typename T>
void printcont(T A) {
  for (auto a : A) {
    cout << a << " ";
  }
  cout << "\n";
}
template <typename T>
T binpowmod(T a, T b, T p) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 0)
    return ((binpowmod(a, b / 2, p) % p) * (binpowmod(a, b / 2, p) % p)) % p;
  return ((a % p) * (binpowmod(a, b - 1, p) % p)) % p;
}
void solve() {
  int n;
  cin >> n;
  long long A[n];
  takeinput(A, n);
  vector<long long> odd_suffix(n, 0);
  if ((n - 1) % 2 == 0) odd_suffix[n - 1] = A[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    if (i % 2 == 0)
      odd_suffix[i] = odd_suffix[i + 1] + A[i];
    else
      odd_suffix[i] = odd_suffix[i + 1];
  }
  odd_suffix.push_back(0);
  long long dp[n + 2][2];
  for (int i = 0; i < n; i++) {
    dp[i][0] = dp[i][1] = 0LL;
  }
  dp[0][0] = A[0];
  dp[0][1] = 0;
  dp[1][0] = A[0];
  dp[1][1] = A[1];
  long long max_ans = A[0] + odd_suffix[1];
  if (n != 1) max_ans = max(max_ans, A[1] + odd_suffix[2]);
  for (int i = 2; i < n; i++) {
    if (i % 2 == 0) {
      dp[i][0] = dp[i - 1][0] + A[i];
      dp[i][1] = max(dp[i - 2][0], dp[i - 2][1]) + A[i - 1];
    } else {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = max(dp[i - 2][0], dp[i - 2][1]) + A[i];
    }
    max_ans = max(max_ans, dp[i][1] + odd_suffix[i + 1]);
  }
  cout << max_ans << "\n";
}
int main() {
  auto start = high_resolution_clock::now();
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<milliseconds>(stop - start);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using namespace std::chrono;
template <typename T>
void printarr(T *A, int n) {
  for (int i = 0; i < n; i++) {
    cout << A[i] << " ";
  }
  cout << "\n";
}
template <typename T>
void arrinit(T *A, int n, T data) {
  for (int i = 0; i < n; i++) {
    A[i] = data;
  }
}
template <typename T>
void takeinput(T *A, int n) {
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
}
template <typename T>
void printcont(T A) {
  for (auto a : A) {
    cout << a << " ";
  }
  cout << "\n";
}
template <typename T>
T binpowmod(T a, T b, T p) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 0)
    return ((binpowmod(a, b / 2, p) % p) * (binpowmod(a, b / 2, p) % p)) % p;
  return ((a % p) * (binpowmod(a, b - 1, p) % p)) % p;
}
void solve() {
  int n;
  cin >> n;
  long long A[n];
  takeinput(A, n);
  vector<long long> odd_suffix(n, 0);
  if ((n - 1) % 2 == 0) odd_suffix[n - 1] = A[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    if (i % 2 == 0)
      odd_suffix[i] = odd_suffix[i + 1] + A[i];
    else
      odd_suffix[i] = odd_suffix[i + 1];
  }
  odd_suffix.push_back(0);
  long long dp[n + 2][2];
  for (int i = 0; i < n; i++) {
    dp[i][0] = dp[i][1] = 0LL;
  }
  dp[0][0] = A[0];
  dp[0][1] = 0;
  dp[1][0] = A[0];
  dp[1][1] = A[1];
  long long max_ans = A[0] + odd_suffix[1];
  if (n != 1) max_ans = max(max_ans, A[1] + odd_suffix[2]);
  for (int i = 2; i < n; i++) {
    if (i % 2 == 0) {
      dp[i][0] = dp[i - 1][0] + A[i];
      dp[i][1] = max(dp[i - 2][0], dp[i - 2][1]) + A[i - 1];
    } else {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = max(dp[i - 2][0], dp[i - 2][1]) + A[i];
    }
    max_ans = max(max_ans, dp[i][1] + odd_suffix[i + 1]);
  }
  cout << max_ans << "\n";
}
int main() {
  auto start = high_resolution_clock::now();
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<milliseconds>(stop - start);
  return 0;
}

```