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
#include <bits/stdc++.h>
using namespace std;
using namespace std;
static const long long INF = 1e15;
static const long long mod = 1e9 + 7;
long long solve(const vector<long long>& as) {
  long long res = -INF;
  long long s = 0;
  for (long long j = 0; j < as.size(); j++) {
    s = max(s + as[j], as[j]);
    res = max(res, s);
  }
  return res;
}
int main() {
  long long t;
  cin >> t;
  for (long long _ = 0; _ < t; _++) {
    long long n;
    cin >> n;
    vector<long long> as(n);
    vector<long long> even((n + 1) / 2);
    vector<long long> odd((n) / 2);
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      cin >> as[i];
      if (i % 2 == 0) {
        even[i / 2] = as[i];
        sum += as[i];
      } else {
        odd[i / 2] = as[i];
      }
    }
    long long u = 0;
    {
      vector<long long> diff(n / 2);
      for (long long i = 0; i < n / 2; i++) {
        diff[i] = odd[i] - even[i];
      }
      long long u1 = solve(diff);
      u = max(u, u1);
    }
    {
      vector<long long> diff(n / 2);
      for (int i = 0; i + 1 < even.size(); i++) {
        diff[i] = odd[i] - even[i + 1];
      }
      long long u1 = solve(diff);
      u = max(u, u1);
    }
    cout << sum + u << endl;
  }
  return 0;
}
