#include <bits/stdc++.h>
using namespace std;
long long d1[4] = {1, -1, 0, 0};
long long d2[4] = {0, 0, 1, -1};
int main() {
  long long T;
  cin >> T;
  for (long long test = (0); test < (long long)(T); test++) {
    long long n;
    cin >> n;
    vector<long long> a(n), d(n / 2);
    long long org = 0;
    for (long long i = (0); i < (long long)(n); i++) {
      cin >> a[i];
      if (i % 2 == 0) org += a[i];
    }
    for (long long i = (0); i < (long long)(n / 2); i++) {
      d[i] = a[2 * i + 1] - a[2 * i];
    }
    long long ans = -1, s = 0;
    for (long long i = (0); i < (long long)(n / 2); i++) {
      s = max(s + d[i], d[i]);
      ans = max(ans, s);
    }
    if (n % 2 == 0) {
      for (long long i = (0); i < (long long)(n / 2 - 1); i++) {
        d[i] = a[2 * i + 1] - a[2 * i + 2];
      }
      long long ans2 = 0, s2 = 0;
      for (long long i = (0); i < (long long)(n / 2 - 1); i++) {
        s2 = max(s2 + d[i], d[i]);
        ans2 = max(ans2, s2);
      }
      ans = max(ans, ans2);
      cout << org + max(0LL, ans) << endl;
    } else {
      for (long long i = (0); i < (long long)(n / 2); i++) {
        d[i] = a[2 * i + 1] - a[2 * i + 2];
      }
      long long ans2 = 0, s2 = 0;
      for (long long i = (0); i < (long long)(n / 2); i++) {
        s2 = max(s2 + d[i], d[i]);
        ans2 = max(ans2, s2);
      }
      ans = max(ans, ans2);
      cout << org + max(0LL, ans) << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
long long d1[4] = {1, -1, 0, 0};
long long d2[4] = {0, 0, 1, -1};
int main() {
  long long T;
  cin >> T;
  for (long long test = (0); test < (long long)(T); test++) {
    long long n;
    cin >> n;
    vector<long long> a(n), d(n / 2);
    long long org = 0;
    for (long long i = (0); i < (long long)(n); i++) {
      cin >> a[i];
      if (i % 2 == 0) org += a[i];
    }
    for (long long i = (0); i < (long long)(n / 2); i++) {
      d[i] = a[2 * i + 1] - a[2 * i];
    }
    long long ans = -1, s = 0;
    for (long long i = (0); i < (long long)(n / 2); i++) {
      s = max(s + d[i], d[i]);
      ans = max(ans, s);
    }
    if (n % 2 == 0) {
      for (long long i = (0); i < (long long)(n / 2 - 1); i++) {
        d[i] = a[2 * i + 1] - a[2 * i + 2];
      }
      long long ans2 = 0, s2 = 0;
      for (long long i = (0); i < (long long)(n / 2 - 1); i++) {
        s2 = max(s2 + d[i], d[i]);
        ans2 = max(ans2, s2);
      }
      ans = max(ans, ans2);
      cout << org + max(0LL, ans) << endl;
    } else {
      for (long long i = (0); i < (long long)(n / 2); i++) {
        d[i] = a[2 * i + 1] - a[2 * i + 2];
      }
      long long ans2 = 0, s2 = 0;
      for (long long i = (0); i < (long long)(n / 2); i++) {
        s2 = max(s2 + d[i], d[i]);
        ans2 = max(ans2, s2);
      }
      ans = max(ans, ans2);
      cout << org + max(0LL, ans) << endl;
    }
  }
  return 0;
}

```