#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 1e9 + 7;
ll pwr(ll a, ll b);
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<ll> b, c;
    for (int i = 1; i < n - 1; i += 2) {
      b.push_back(a[i] - a[i + 1]);
    }
    for (int i = 0; i < n - 1; i += 2) {
      c.push_back(a[i + 1] - a[i]);
    }
    int m = b.size();
    ll best = 0, sum = 0;
    for (int i = 0; i < m; i++) {
      sum = max(sum + b[i], b[i]);
      best = max(best, sum);
    }
    sum = 0;
    for (int i = 0; i < (int)c.size(); i++) {
      sum = max(sum + c[i], c[i]);
      best = max(best, sum);
    }
    ll ans = 0;
    for (int i = 0; i < n; i += 2) {
      ans += a[i];
    }
    if (best > 0) {
      cout << ans + best << "\n";
    } else {
      cout << ans << "\n";
    }
  }
}
ll pwr(ll a, ll b) {
  a %= mod;
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}

### Prompt

Your challenge is to write 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;
using ll = long long;
const int mod = 1e9 + 7;
ll pwr(ll a, ll b);
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<ll> b, c;
    for (int i = 1; i < n - 1; i += 2) {
      b.push_back(a[i] - a[i + 1]);
    }
    for (int i = 0; i < n - 1; i += 2) {
      c.push_back(a[i + 1] - a[i]);
    }
    int m = b.size();
    ll best = 0, sum = 0;
    for (int i = 0; i < m; i++) {
      sum = max(sum + b[i], b[i]);
      best = max(best, sum);
    }
    sum = 0;
    for (int i = 0; i < (int)c.size(); i++) {
      sum = max(sum + c[i], c[i]);
      best = max(best, sum);
    }
    ll ans = 0;
    for (int i = 0; i < n; i += 2) {
      ans += a[i];
    }
    if (best > 0) {
      cout << ans + best << "\n";
    } else {
      cout << ans << "\n";
    }
  }
}
ll pwr(ll a, ll b) {
  a %= mod;
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}

```