#include <bits/stdc++.h>
using namespace std;
long long denom = 1e9 + 7;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= denom;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % denom;
    a = a * a % denom;
  }
  return res;
}
long long maxSub(vector<long long>& v) {
  long long mn = 0;
  long long sum = 0;
  long long result = 0;
  for (long long i = (0); i < (v.size()); i++) {
    sum += v[i];
    result = max(result, sum - mn);
    mn = min(mn, sum);
  }
  return result;
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  long long t;
  cin >> t;
  for (long long i = (0); i < (t); i++) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long j = (0); j < (n); j++) cin >> a[j];
    long long sum = 0;
    for (long long j = (0); j < (n); j++) sum += j % 2 == 0 ? a[j] : 0;
    vector<long long> v1;
    for (long long j = 1; j < n; j += 2) v1.push_back(a[j] - a[j - 1]);
    vector<long long> v2;
    for (long long j = 1; j < n - 1; j += 2) v2.push_back(a[j] - a[j + 1]);
    long long delta = max(maxSub(v1), maxSub(v2));
    sum += delta;
    cout << sum << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 denom = 1e9 + 7;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= denom;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % denom;
    a = a * a % denom;
  }
  return res;
}
long long maxSub(vector<long long>& v) {
  long long mn = 0;
  long long sum = 0;
  long long result = 0;
  for (long long i = (0); i < (v.size()); i++) {
    sum += v[i];
    result = max(result, sum - mn);
    mn = min(mn, sum);
  }
  return result;
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  long long t;
  cin >> t;
  for (long long i = (0); i < (t); i++) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long j = (0); j < (n); j++) cin >> a[j];
    long long sum = 0;
    for (long long j = (0); j < (n); j++) sum += j % 2 == 0 ? a[j] : 0;
    vector<long long> v1;
    for (long long j = 1; j < n; j += 2) v1.push_back(a[j] - a[j - 1]);
    vector<long long> v2;
    for (long long j = 1; j < n - 1; j += 2) v2.push_back(a[j] - a[j + 1]);
    long long delta = max(maxSub(v1), maxSub(v2));
    sum += delta;
    cout << sum << endl;
  }
  return 0;
}

```