#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const long long INF = 1e18;
long long f(vector<long long> b) {
  long long mini = 0;
  long long ret = 0;
  long long now = 0;
  for (int i = 0; i < b.size(); i++) {
    now += b[i];
    chmax(ret, now - mini);
    chmin(mini, now);
  }
  return ret;
}
void solve() {
  long long N;
  cin >> N;
  vector<long long> a(N);
  long long sum = 0;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    if (i % 2 == 0) sum += a[i];
  }
  vector<long long> b;
  for (int i = 0; i + 1 < N; i += 2) {
    b.push_back(a[i + 1] - a[i]);
  }
  long long tmp = f(b);
  b.clear();
  for (int i = 1; i + 1 < N; i += 2) {
    b.push_back(a[i] - a[i + 1]);
  }
  chmax(tmp, f(b));
  long long ans = sum + tmp;
  cout << ans << "\n";
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long T;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Generate 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;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const long long INF = 1e18;
long long f(vector<long long> b) {
  long long mini = 0;
  long long ret = 0;
  long long now = 0;
  for (int i = 0; i < b.size(); i++) {
    now += b[i];
    chmax(ret, now - mini);
    chmin(mini, now);
  }
  return ret;
}
void solve() {
  long long N;
  cin >> N;
  vector<long long> a(N);
  long long sum = 0;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    if (i % 2 == 0) sum += a[i];
  }
  vector<long long> b;
  for (int i = 0; i + 1 < N; i += 2) {
    b.push_back(a[i + 1] - a[i]);
  }
  long long tmp = f(b);
  b.clear();
  for (int i = 1; i + 1 < N; i += 2) {
    b.push_back(a[i] - a[i + 1]);
  }
  chmax(tmp, f(b));
  long long ans = sum + tmp;
  cout << ans << "\n";
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```