#include <bits/stdc++.h>
using namespace std;
vector<int> adj[200105], new_adj[200105];
int xMove[] = {0, 1, 0, -1, 1, 1, -1, -1, 2, 2, -2, -2, 1, 1, -1, -1};
int yMove[] = {1, 0, -1, 0, 1, -1, 1, -1, 1, -1, 1, -1, 2, -2, 2, -2};
void clr(int n) {
  for (int i = 0; i < n + 2; i++) adj[i].clear();
}
int main() {
  cin.tie(0), cout.tie(0);
  ios_base::sync_with_stdio(0);
  int TC = 1;
  int n, q, m, e, t, y, c, f, k;
  cin >> TC;
  for (int t = 1; t <= TC; t++) {
    cin >> n;
    long long int arr[n];
    long long int ans = 0;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      if (i % 2 == 0) ans += arr[i];
    }
    vector<long long int> vec1, vec2;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        if (i + 1 < n) vec1.push_back(arr[i + 1] - arr[i]);
      } else {
        if (i + 1 < n) vec2.push_back(arr[i] - arr[i + 1]);
      }
    }
    long long int cur = 0, mx = 0;
    for (long long int v : vec1) {
      cur += v;
      mx = max(mx, cur);
      if (cur < 0) cur = 0;
    }
    cur = 0;
    for (long long int v : vec2) {
      cur += v;
      mx = max(mx, cur);
      if (cur < 0) cur = 0;
    }
    cout << ans + mx << endl;
  }
  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;
vector<int> adj[200105], new_adj[200105];
int xMove[] = {0, 1, 0, -1, 1, 1, -1, -1, 2, 2, -2, -2, 1, 1, -1, -1};
int yMove[] = {1, 0, -1, 0, 1, -1, 1, -1, 1, -1, 1, -1, 2, -2, 2, -2};
void clr(int n) {
  for (int i = 0; i < n + 2; i++) adj[i].clear();
}
int main() {
  cin.tie(0), cout.tie(0);
  ios_base::sync_with_stdio(0);
  int TC = 1;
  int n, q, m, e, t, y, c, f, k;
  cin >> TC;
  for (int t = 1; t <= TC; t++) {
    cin >> n;
    long long int arr[n];
    long long int ans = 0;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      if (i % 2 == 0) ans += arr[i];
    }
    vector<long long int> vec1, vec2;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        if (i + 1 < n) vec1.push_back(arr[i + 1] - arr[i]);
      } else {
        if (i + 1 < n) vec2.push_back(arr[i] - arr[i + 1]);
      }
    }
    long long int cur = 0, mx = 0;
    for (long long int v : vec1) {
      cur += v;
      mx = max(mx, cur);
      if (cur < 0) cur = 0;
    }
    cur = 0;
    for (long long int v : vec2) {
      cur += v;
      mx = max(mx, cur);
      if (cur < 0) cur = 0;
    }
    cout << ans + mx << endl;
  }
  return 0;
}

```