#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long kadane(vector<long long> a, long long size) {
  long long max_so_far = INT_MIN, max_ending_here = 0;
  for (long long i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
void solve() {
  long long n, k, m = 0, i = 0, j = 0, l, r, x, y;
  cin >> n;
  vector<long long> a(n, 0), b(n, 0), c(n, 0);
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 0) m += a[i];
  }
  int t = 0;
  int p = 0;
  for (i = 1; i < n - 1; i += 2) b[t] = a[i] - a[i + 1], t++;
  for (i = 0; i < n - 1; i += 2) c[p] = a[i + 1] - a[i], p++;
  cout << m + max({kadane(c, p), kadane(b, t), 0LL}) << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) 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;
const int N = 200005;
long long kadane(vector<long long> a, long long size) {
  long long max_so_far = INT_MIN, max_ending_here = 0;
  for (long long i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
void solve() {
  long long n, k, m = 0, i = 0, j = 0, l, r, x, y;
  cin >> n;
  vector<long long> a(n, 0), b(n, 0), c(n, 0);
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 0) m += a[i];
  }
  int t = 0;
  int p = 0;
  for (i = 1; i < n - 1; i += 2) b[t] = a[i] - a[i + 1], t++;
  for (i = 0; i < n - 1; i += 2) c[p] = a[i + 1] - a[i], p++;
  cout << m + max({kadane(c, p), kadane(b, t), 0LL}) << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```