#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
int ll_max = 1e6;
vector<long long int> fact;
void factorial(long long int n) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= M;
  }
}
long long int binpow(long long int val, long long int deg) {
  if (!deg) return 1 % M;
  if (deg & 1) return binpow(val, deg - 1) * val % M;
  long long int res = binpow(val, deg >> 1);
  return (res * res) % M;
}
long long int combination(long long int n, long long int i) {
  long long int res = fact[n];
  long long int div = fact[n - i] * fact[i];
  div %= M;
  div = binpow(div, M - 2);
  return (res * div) % M;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, m, i;
  long long int j, t, x, y, z, k, d;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n];
    long long int sum = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) sum += a[i];
    }
    vector<long long int> t1, t2;
    for (i = 0; i < n; i += 2) {
      if (i + 1 < n) t1.push_back(a[i + 1] - a[i]);
    }
    for (i = 1; i < n; i += 2) {
      if (i + 1 < n) t2.push_back(a[i] - a[i + 1]);
    }
    long long int max = 0;
    long long int max_so_far = 0, max_ending_here = 0;
    for (i = 0; i < t1.size(); i++) {
      max_ending_here = max_ending_here + t1[i];
      if (max_so_far < max_ending_here) max_so_far = max_ending_here;
      if (max_ending_here < 0) max_ending_here = 0;
    }
    max = max_so_far;
    max_so_far = 0, max_ending_here = 0;
    for (i = 0; i < t2.size(); i++) {
      max_ending_here = max_ending_here + t2[i];
      if (max_so_far < max_ending_here) max_so_far = max_ending_here;
      if (max_ending_here < 0) max_ending_here = 0;
    }
    if (max < max_so_far) max = max_so_far;
    cout << max + sum << '\n';
  }
  return 0;
}

### 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 int M = 1e9 + 7;
int ll_max = 1e6;
vector<long long int> fact;
void factorial(long long int n) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= M;
  }
}
long long int binpow(long long int val, long long int deg) {
  if (!deg) return 1 % M;
  if (deg & 1) return binpow(val, deg - 1) * val % M;
  long long int res = binpow(val, deg >> 1);
  return (res * res) % M;
}
long long int combination(long long int n, long long int i) {
  long long int res = fact[n];
  long long int div = fact[n - i] * fact[i];
  div %= M;
  div = binpow(div, M - 2);
  return (res * div) % M;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, m, i;
  long long int j, t, x, y, z, k, d;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n];
    long long int sum = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) sum += a[i];
    }
    vector<long long int> t1, t2;
    for (i = 0; i < n; i += 2) {
      if (i + 1 < n) t1.push_back(a[i + 1] - a[i]);
    }
    for (i = 1; i < n; i += 2) {
      if (i + 1 < n) t2.push_back(a[i] - a[i + 1]);
    }
    long long int max = 0;
    long long int max_so_far = 0, max_ending_here = 0;
    for (i = 0; i < t1.size(); i++) {
      max_ending_here = max_ending_here + t1[i];
      if (max_so_far < max_ending_here) max_so_far = max_ending_here;
      if (max_ending_here < 0) max_ending_here = 0;
    }
    max = max_so_far;
    max_so_far = 0, max_ending_here = 0;
    for (i = 0; i < t2.size(); i++) {
      max_ending_here = max_ending_here + t2[i];
      if (max_so_far < max_ending_here) max_so_far = max_ending_here;
      if (max_ending_here < 0) max_ending_here = 0;
    }
    if (max < max_so_far) max = max_so_far;
    cout << max + sum << '\n';
  }
  return 0;
}

```