#include <bits/stdc++.h>
using namespace std;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
void sanjeev(int kkkk) {
  int t;
  cin >> t;
  while (t > 0) {
    t--;
    int n;
    cin >> n;
    vector<int> A;
    long long int a = 0;
    for (int i = 0; i < n; i++) {
      int tmp;
      cin >> tmp;
      A.push_back(tmp);
      if (i % 2 == 0) a += tmp;
    }
    long long int b = 0, m = 0, m2 = 0;
    for (int i = 0; i < n - 1; i += 2) {
      b += (A[i + 1] - A[i]);
      m = max(b, m);
      if (b < 0) b = 0;
    }
    b = 0;
    for (int i = 1; i < n - 1; i += 2) {
      b += (A[i] - A[i + 1]);
      m2 = max(m2, b);
      if (b < 0) b = 0;
    }
    a += max(m, m2);
    cout << a << '\n';
  }
}
int main() {
  if (!0) ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    sanjeev(i + 1);
  }
  if (0) system("pause");
}

### Prompt

Your challenge is to write a cpp solution to 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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
void sanjeev(int kkkk) {
  int t;
  cin >> t;
  while (t > 0) {
    t--;
    int n;
    cin >> n;
    vector<int> A;
    long long int a = 0;
    for (int i = 0; i < n; i++) {
      int tmp;
      cin >> tmp;
      A.push_back(tmp);
      if (i % 2 == 0) a += tmp;
    }
    long long int b = 0, m = 0, m2 = 0;
    for (int i = 0; i < n - 1; i += 2) {
      b += (A[i + 1] - A[i]);
      m = max(b, m);
      if (b < 0) b = 0;
    }
    b = 0;
    for (int i = 1; i < n - 1; i += 2) {
      b += (A[i] - A[i + 1]);
      m2 = max(m2, b);
      if (b < 0) b = 0;
    }
    a += max(m, m2);
    cout << a << '\n';
  }
}
int main() {
  if (!0) ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    sanjeev(i + 1);
  }
  if (0) system("pause");
}

```