#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
const int INF = 0x3f3f3f3f, N = 2e5 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 7;
int n, a[N];
ll dp[N][2];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  ll sum = 0, ext = 0;
  for (int i = 0; i < n; i += 2) sum += a[i];
  for (int i = 0; i + 1 < n; i++) {
    if (i & 1) {
      dp[i][1] = a[i] - a[i + 1];
      if (i >= 2 && dp[i - 2][1] > 0) dp[i][1] += dp[i - 2][1];
      ext = max(ext, dp[i][1]);
    } else {
      dp[i][0] = a[i + 1] - a[i];
      if (i >= 2 && dp[i - 2][0] > 0) dp[i][0] += dp[i - 2][0];
      ext = max(ext, dp[i][0]);
    }
  }
  cout << sum + ext << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int _;
  cin >> _;
  while (_--) solve();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
const int INF = 0x3f3f3f3f, N = 2e5 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 7;
int n, a[N];
ll dp[N][2];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  ll sum = 0, ext = 0;
  for (int i = 0; i < n; i += 2) sum += a[i];
  for (int i = 0; i + 1 < n; i++) {
    if (i & 1) {
      dp[i][1] = a[i] - a[i + 1];
      if (i >= 2 && dp[i - 2][1] > 0) dp[i][1] += dp[i - 2][1];
      ext = max(ext, dp[i][1]);
    } else {
      dp[i][0] = a[i + 1] - a[i];
      if (i >= 2 && dp[i - 2][0] > 0) dp[i][0] += dp[i - 2][0];
      ext = max(ext, dp[i][0]);
    }
  }
  cout << sum + ext << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int _;
  cin >> _;
  while (_--) solve();
  return 0;
}

```