#include <bits/stdc++.h>
using namespace std;
const long long LINF = 1e18 + 100100, INF = 2e9 + 100100;
const double EPS = 1e-9, PI = acos(-1);
const long long MOD = 1e9 + 7, P = 239;
const long long MOD1 = 1e9 + 9, P1 = 117;
const long long MAXN = 1e7, MAXLOG = 21;
inline void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n), pref(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    if (i == 0) {
      pref[i] = -a[i];
    } else {
      pref[i] = pref[i - 1];
      if (i & 1) {
        pref[i] += a[i];
      } else {
        pref[i] -= a[i];
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; i += 2) {
    ans += a[i];
  }
  long long saveAns = ans;
  vector<long long> last(2);
  for (long long i = 0; i < n; ++i) {
    last[i % 2] = min(last[i % 2], pref[i]);
    ans = max(ans, saveAns + pref[i] - last[i % 2]);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
const long long LINF = 1e18 + 100100, INF = 2e9 + 100100;
const double EPS = 1e-9, PI = acos(-1);
const long long MOD = 1e9 + 7, P = 239;
const long long MOD1 = 1e9 + 9, P1 = 117;
const long long MAXN = 1e7, MAXLOG = 21;
inline void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n), pref(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    if (i == 0) {
      pref[i] = -a[i];
    } else {
      pref[i] = pref[i - 1];
      if (i & 1) {
        pref[i] += a[i];
      } else {
        pref[i] -= a[i];
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; i += 2) {
    ans += a[i];
  }
  long long saveAns = ans;
  vector<long long> last(2);
  for (long long i = 0; i < n; ++i) {
    last[i % 2] = min(last[i % 2], pref[i]);
    ans = max(ans, saveAns + pref[i] - last[i % 2]);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}

```