#include <bits/stdc++.h>
using namespace std;
void DBG() { cerr << "]\n"; }
template <typename T, typename... Args>
void DBG(T first, Args... args) {
  cerr << first;
  if (sizeof...(args)) cerr << ", ";
  DBG(args...);
}
long long int del_x[4] = {-1, 1, 0, 0};
long long int del_y[4] = {0, 0, 1, -1};
long long int XX[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
long long int YY[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < (n); i++) cin >> a[i];
  long long int curr_sum = 0;
  long long int max_so_far = 0;
  for (long long int i = 0; i < n - n % 2; i += 2) {
    long long int extra = a[i + 1] - a[i];
    curr_sum = max(0ll, curr_sum + extra);
    max_so_far = max(curr_sum, max_so_far);
  }
  curr_sum = 0;
  for (long long int i = 1; i < n - (n % 2 == 0); i += 2) {
    long long int extra = a[i] - a[i + 1];
    curr_sum = max(0ll, curr_sum + extra);
    max_so_far = max(curr_sum, max_so_far);
  }
  long long int sum = 0;
  for (long long int i = 0; i < (n); i++) {
    if (i % 2 == 0) sum += a[i];
  }
  cout << sum + max_so_far << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t = 1;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    solve();
  }
}

### 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;
void DBG() { cerr << "]\n"; }
template <typename T, typename... Args>
void DBG(T first, Args... args) {
  cerr << first;
  if (sizeof...(args)) cerr << ", ";
  DBG(args...);
}
long long int del_x[4] = {-1, 1, 0, 0};
long long int del_y[4] = {0, 0, 1, -1};
long long int XX[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
long long int YY[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < (n); i++) cin >> a[i];
  long long int curr_sum = 0;
  long long int max_so_far = 0;
  for (long long int i = 0; i < n - n % 2; i += 2) {
    long long int extra = a[i + 1] - a[i];
    curr_sum = max(0ll, curr_sum + extra);
    max_so_far = max(curr_sum, max_so_far);
  }
  curr_sum = 0;
  for (long long int i = 1; i < n - (n % 2 == 0); i += 2) {
    long long int extra = a[i] - a[i + 1];
    curr_sum = max(0ll, curr_sum + extra);
    max_so_far = max(curr_sum, max_so_far);
  }
  long long int sum = 0;
  for (long long int i = 0; i < (n); i++) {
    if (i % 2 == 0) sum += a[i];
  }
  cout << sum + max_so_far << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t = 1;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    solve();
  }
}

```