#include <bits/stdc++.h>
using namespace std;
const int64_t INF = 0x3f3f3f3f;
const int64_t INFLL = 0x3f3f3f3f3f3f3f3f;
const int64_t MOD = 1e9 + 7;
const double EPS = 1e-9;
const double PI = acos(-1);
const int64_t MAX = 2e5 + 10;
const int64_t NMAX = 2e5 + 10;
const int64_t MMAX = 2e5 + 10;
int64_t n;
int64_t arr[NMAX];
int64_t mem[NMAX][3][2];
int64_t vis[NMAX][3][2], vid;
int64_t go(int64_t i, int64_t k, int64_t p) {
  if (!p)
    if ((i == n - 1) && (k == 1)) return -INF;
  if (p)
    if ((i >= n) && (k == 1)) return -INF;
  if (i >= n) return 0;
  if (vis[i][k][p] == vid) return mem[i][k][p];
  vis[i][k][p] = vid;
  int64_t ans = 0;
  if (k == 0) {
    int64_t stPath = arr[i] + go(i + 2, 0, 0);
    int64_t ndPath = arr[i] + go(i + 1, 1, 0);
    int64_t rdPath = arr[i] + go(i + 3, 1, 1);
    ans = max(stPath, max(ndPath, rdPath));
  } else if (k == 1) {
    int64_t stPath = arr[i] + go(i + 2, 1, p);
    int64_t ndPath = arr[i] + go(i + 1 + (!p * 2), 2, 0);
    ans = max(stPath, ndPath);
  } else {
    int64_t stPath = arr[i] + go(i + 2, 2, 0);
    ans = stPath;
  }
  return mem[i][k][p] = ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int64_t t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int64_t i = 0; i < n; i++) cin >> arr[i];
    vid++;
    cout << max(go(0, 0, 0), go(1, 1, 1)) << '\n';
  }
}

### 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;
const int64_t INF = 0x3f3f3f3f;
const int64_t INFLL = 0x3f3f3f3f3f3f3f3f;
const int64_t MOD = 1e9 + 7;
const double EPS = 1e-9;
const double PI = acos(-1);
const int64_t MAX = 2e5 + 10;
const int64_t NMAX = 2e5 + 10;
const int64_t MMAX = 2e5 + 10;
int64_t n;
int64_t arr[NMAX];
int64_t mem[NMAX][3][2];
int64_t vis[NMAX][3][2], vid;
int64_t go(int64_t i, int64_t k, int64_t p) {
  if (!p)
    if ((i == n - 1) && (k == 1)) return -INF;
  if (p)
    if ((i >= n) && (k == 1)) return -INF;
  if (i >= n) return 0;
  if (vis[i][k][p] == vid) return mem[i][k][p];
  vis[i][k][p] = vid;
  int64_t ans = 0;
  if (k == 0) {
    int64_t stPath = arr[i] + go(i + 2, 0, 0);
    int64_t ndPath = arr[i] + go(i + 1, 1, 0);
    int64_t rdPath = arr[i] + go(i + 3, 1, 1);
    ans = max(stPath, max(ndPath, rdPath));
  } else if (k == 1) {
    int64_t stPath = arr[i] + go(i + 2, 1, p);
    int64_t ndPath = arr[i] + go(i + 1 + (!p * 2), 2, 0);
    ans = max(stPath, ndPath);
  } else {
    int64_t stPath = arr[i] + go(i + 2, 2, 0);
    ans = stPath;
  }
  return mem[i][k][p] = ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int64_t t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int64_t i = 0; i < n; i++) cin >> arr[i];
    vid++;
    cout << max(go(0, 0, 0), go(1, 1, 1)) << '\n';
  }
}

```