#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long power(long long x, long long p) {
  if (p == 0) return 1;
  if (p == 1) return x;
  if (p % 2) return x * power(x, p - 1);
  long long y = power(x, p / 2);
  return y * y;
}
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  pair<long long, long long> modd, meven;
  long long odd = 0, even = 0, l = -1, r = -1, res = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2)
      odd += arr[i];
    else
      even += arr[i];
    long long cur = odd - even;
    pair<long long, long long> p;
    if (i % 2)
      p = meven;
    else
      p = modd;
    long long val = cur - p.first;
    if (val > res) {
      l = p.second;
      r = i;
      res = val;
    }
    long long x = odd - even + arr[i] * (i % 2 ? -1 : 1);
    if (i % 2 == 0) {
      if (x < meven.first) meven = make_pair(x, i);
    } else {
      if (x < modd.first) modd = make_pair(x, i);
    }
  }
  if (l != -1) {
    for (int i = 0; i < (r - l + 1) / 2; i++) {
      swap(arr[l + i], arr[r - i]);
    }
  }
  odd = 0;
  even = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2)
      odd += arr[i];
    else
      even += arr[i];
  }
  cout << max(odd, even) << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long power(long long x, long long p) {
  if (p == 0) return 1;
  if (p == 1) return x;
  if (p % 2) return x * power(x, p - 1);
  long long y = power(x, p / 2);
  return y * y;
}
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  pair<long long, long long> modd, meven;
  long long odd = 0, even = 0, l = -1, r = -1, res = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2)
      odd += arr[i];
    else
      even += arr[i];
    long long cur = odd - even;
    pair<long long, long long> p;
    if (i % 2)
      p = meven;
    else
      p = modd;
    long long val = cur - p.first;
    if (val > res) {
      l = p.second;
      r = i;
      res = val;
    }
    long long x = odd - even + arr[i] * (i % 2 ? -1 : 1);
    if (i % 2 == 0) {
      if (x < meven.first) meven = make_pair(x, i);
    } else {
      if (x < modd.first) modd = make_pair(x, i);
    }
  }
  if (l != -1) {
    for (int i = 0; i < (r - l + 1) / 2; i++) {
      swap(arr[l + i], arr[r - i]);
    }
  }
  odd = 0;
  even = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2)
      odd += arr[i];
    else
      even += arr[i];
  }
  cout << max(odd, even) << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```