#include <bits/stdc++.h>
using namespace std;
bool com(pair<int, int> x, pair<int, int> y) {
  if (x.first == y.first) return x.second < y.second;
  return x.first < y.first;
}
long long int power(long long int x, long long int y) {
  long long int prod = 1;
  while (y) {
    if (y & 1) prod = (prod * x) % 1000000007;
    x = (x * x) % 1000000007;
    y /= 2;
  }
  return prod;
}
const int N = 2e5 + 7;
void solve() {
  int n;
  cin >> n;
  long long int a[n + 1];
  long long int sum = 0, ans = 0;
  vector<long long int> od, ev;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i += 2) sum += a[i];
  for (int i = 1; i < n; i += 2) {
    ev.push_back(a[i + 1] - a[i]);
  }
  ans = sum;
  long long int x = 0;
  for (int i = 2; i < n; i += 2) od.push_back(a[i] - a[i + 1]);
  for (int i = 0; i < od.size(); i++) {
    x = max(x + od[i], od[i]);
    ans = max(ans, sum + x);
  }
  x = 0;
  for (int i = 0; i < ev.size(); i++) {
    x = max(x + ev[i], ev[i]);
    ans = max(ans, sum + x);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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;
bool com(pair<int, int> x, pair<int, int> y) {
  if (x.first == y.first) return x.second < y.second;
  return x.first < y.first;
}
long long int power(long long int x, long long int y) {
  long long int prod = 1;
  while (y) {
    if (y & 1) prod = (prod * x) % 1000000007;
    x = (x * x) % 1000000007;
    y /= 2;
  }
  return prod;
}
const int N = 2e5 + 7;
void solve() {
  int n;
  cin >> n;
  long long int a[n + 1];
  long long int sum = 0, ans = 0;
  vector<long long int> od, ev;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i += 2) sum += a[i];
  for (int i = 1; i < n; i += 2) {
    ev.push_back(a[i + 1] - a[i]);
  }
  ans = sum;
  long long int x = 0;
  for (int i = 2; i < n; i += 2) od.push_back(a[i] - a[i + 1]);
  for (int i = 0; i < od.size(); i++) {
    x = max(x + od[i], od[i]);
    ans = max(ans, sum + x);
  }
  x = 0;
  for (int i = 0; i < ev.size(); i++) {
    x = max(x + ev[i], ev[i]);
    ans = max(ans, sum + x);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```