#include <bits/stdc++.h>
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
long long dx[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long dy[] = {1, 1, 0, -1, -1, -1, 0, 1};
void print(vector<long long> &v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    long long sum = 0;
    vector<long long> diff(n + 1);
    for (i = 1; i <= n; i++) {
      long long in;
      cin >> in;
      if (i % 2 == 0) {
        diff[i] -= in;
      } else {
        diff[i] += in;
        sum += in;
      }
      diff[i] += diff[i - 1];
    }
    long long minodd = 0, mineven = 0;
    long long ans = sum;
    for (i = 1; i <= n; i++) {
      if (i % 2 == 0) {
        ans = max(ans, sum - (diff[i] - minodd));
        minodd = max(minodd, diff[i]);
      } else {
        ans = max(ans, sum - (diff[i] - mineven));
        mineven = max(mineven, diff[i]);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 inf = 0x3f3f3f3f3f3f3f3f;
long long dx[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long dy[] = {1, 1, 0, -1, -1, -1, 0, 1};
void print(vector<long long> &v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    long long sum = 0;
    vector<long long> diff(n + 1);
    for (i = 1; i <= n; i++) {
      long long in;
      cin >> in;
      if (i % 2 == 0) {
        diff[i] -= in;
      } else {
        diff[i] += in;
        sum += in;
      }
      diff[i] += diff[i - 1];
    }
    long long minodd = 0, mineven = 0;
    long long ans = sum;
    for (i = 1; i <= n; i++) {
      if (i % 2 == 0) {
        ans = max(ans, sum - (diff[i] - minodd));
        minodd = max(minodd, diff[i]);
      } else {
        ans = max(ans, sum - (diff[i] - mineven));
        mineven = max(mineven, diff[i]);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```