#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); };
int dr[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dc[] = {1, 0, -1, 0, 1, -1, 1, -1};
int n, m, k, t;
int a[200001];
int main() {
  ios::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  cin >> t;
  while (t--) {
    cin >> n;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) ans += a[i];
    }
    long long mx = 0;
    long long tmp = 0;
    for (int i = 0; i < n / 2 * 2; i += 2) {
      tmp -= a[i];
      tmp += a[i + 1];
      if (tmp < 0) {
        tmp = 0;
      } else {
        mx = max(mx, tmp);
      }
    }
    tmp = 0;
    for (int i = 1; i < n - 1; i += 2) {
      tmp += a[i];
      tmp -= a[i + 1];
      if (tmp < 0) {
        tmp = 0;
      } else {
        mx = max(mx, tmp);
      }
    }
    cout << ans + mx << '\n';
  }
  return 0;
}

### 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;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); };
int dr[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dc[] = {1, 0, -1, 0, 1, -1, 1, -1};
int n, m, k, t;
int a[200001];
int main() {
  ios::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  cin >> t;
  while (t--) {
    cin >> n;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) ans += a[i];
    }
    long long mx = 0;
    long long tmp = 0;
    for (int i = 0; i < n / 2 * 2; i += 2) {
      tmp -= a[i];
      tmp += a[i + 1];
      if (tmp < 0) {
        tmp = 0;
      } else {
        mx = max(mx, tmp);
      }
    }
    tmp = 0;
    for (int i = 1; i < n - 1; i += 2) {
      tmp += a[i];
      tmp -= a[i + 1];
      if (tmp < 0) {
        tmp = 0;
      } else {
        mx = max(mx, tmp);
      }
    }
    cout << ans + mx << '\n';
  }
  return 0;
}

```