#include <bits/stdc++.h>
using namespace std;
const long long N = 1500007, NS = 2e3 + 7, lg = 18, sq = 550, inf = 1e9 + 7,
                mod = 1e9 + 7, SM = 101;
const long double eps = 1e-9;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long _;
  cin >> _;
  while (_--) {
    long long n;
    cin >> n;
    vector<long long> a(n), s(n + 1);
    for (long long i = 0; i < n; i++) cin >> a[i];
    for (long long i = n - 1; i >= 0; i--) {
      s[i] = s[i + 1];
      if (i % 2 == 0)
        s[i] += a[i];
      else
        s[i] -= a[i];
    }
    long long c = 0, ans = s[0], sum = 0, c1 = 0, c0 = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      if (i % 2 == 0) {
        c += a[i];
        ans = max(ans, 2 * c0 - c + s[i + 1]);
        c0 = max(c0, c);
      } else {
        c -= a[i];
        ans = max(ans, 2 * c1 - c + s[i + 1]);
        c1 = max(c1, c);
      }
    }
    cout << ans + (sum - ans) / 2 << '\n';
  }
}

### 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 long long N = 1500007, NS = 2e3 + 7, lg = 18, sq = 550, inf = 1e9 + 7,
                mod = 1e9 + 7, SM = 101;
const long double eps = 1e-9;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long _;
  cin >> _;
  while (_--) {
    long long n;
    cin >> n;
    vector<long long> a(n), s(n + 1);
    for (long long i = 0; i < n; i++) cin >> a[i];
    for (long long i = n - 1; i >= 0; i--) {
      s[i] = s[i + 1];
      if (i % 2 == 0)
        s[i] += a[i];
      else
        s[i] -= a[i];
    }
    long long c = 0, ans = s[0], sum = 0, c1 = 0, c0 = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      if (i % 2 == 0) {
        c += a[i];
        ans = max(ans, 2 * c0 - c + s[i + 1]);
        c0 = max(c0, c);
      } else {
        c -= a[i];
        ans = max(ans, 2 * c1 - c + s[i + 1]);
        c1 = max(c1, c);
      }
    }
    cout << ans + (sum - ans) / 2 << '\n';
  }
}

```