#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
const long long inf = 1e18;
long long i, j, k, n, m, x, y, ma = -inf, mi = inf, qq = 1, w, fr, to, t;
char ch;
long long a[200001], p0[200001], p1[200001], p[200001];
void out() {
  for (i = 1; i <= n; i++) cout << p[i] << '\t';
  cout << '\n';
  for (i = 1; i <= n; i++) cout << p0[i] << '\t';
  cout << '\n';
  for (i = 1; i <= n; i++) cout << p1[i] << '\t';
  cout << '\n';
}
void solve() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (i % 2 == 1) {
      p0[i] = a[i];
      if (i - 2 > -1) p0[i] += p0[i - 2];
      p1[i] = p1[i - 1];
    } else {
      p1[i] = a[i];
      if (i - 2 > -1) p1[i] += p1[i - 2];
      p0[i] = p0[i - 1];
    }
  }
  for (i = 1; i <= n; i++) {
    p[i] = p1[i] - p0[i];
  }
  set<long long> s1, s0;
  s1.insert(0);
  long long ans = p0[n];
  for (i = 1; i <= n; i += 2) {
    if (!s0.empty()) ans = max(ans, p0[n] + p[i] - *s0.begin());
    s0.insert(p[i]);
  }
  for (i = 2; i <= n; i += 2) {
    ans = max(ans, p0[n] + p[i] - *s1.begin());
    s1.insert(p[i]);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> qq;
  while (qq--) {
    solve();
  }
}

### Prompt

In CPP, your task is to solve 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;
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
const long long inf = 1e18;
long long i, j, k, n, m, x, y, ma = -inf, mi = inf, qq = 1, w, fr, to, t;
char ch;
long long a[200001], p0[200001], p1[200001], p[200001];
void out() {
  for (i = 1; i <= n; i++) cout << p[i] << '\t';
  cout << '\n';
  for (i = 1; i <= n; i++) cout << p0[i] << '\t';
  cout << '\n';
  for (i = 1; i <= n; i++) cout << p1[i] << '\t';
  cout << '\n';
}
void solve() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (i % 2 == 1) {
      p0[i] = a[i];
      if (i - 2 > -1) p0[i] += p0[i - 2];
      p1[i] = p1[i - 1];
    } else {
      p1[i] = a[i];
      if (i - 2 > -1) p1[i] += p1[i - 2];
      p0[i] = p0[i - 1];
    }
  }
  for (i = 1; i <= n; i++) {
    p[i] = p1[i] - p0[i];
  }
  set<long long> s1, s0;
  s1.insert(0);
  long long ans = p0[n];
  for (i = 1; i <= n; i += 2) {
    if (!s0.empty()) ans = max(ans, p0[n] + p[i] - *s0.begin());
    s0.insert(p[i]);
  }
  for (i = 2; i <= n; i += 2) {
    ans = max(ans, p0[n] + p[i] - *s1.begin());
    s1.insert(p[i]);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> qq;
  while (qq--) {
    solve();
  }
}

```