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;
template <typename T, typename U>
T &ctmax(T &x, const U &y) {
  return x = max<T>(x, y);
}
template <typename T, typename U>
T &ctmin(T &x, const U &y) {
  return x = min<T>(x, y);
}
int SOLVE() {
  int n;
  cin >> n;
  vector<int> a(n);
  long long tot = 0;
  for (auto i = 0; i < n; ++i) {
    cin >> a[i];
    if (!(i & 1)) {
      tot += a[i];
    }
  }
  vector<long long> pref(n + 1);
  for (auto i = 0; i < n; ++i) {
    pref[i + 1] = pref[i] + (i & 1 ? a[i] : -a[i]);
  }
  array<long long, 2> mval{numeric_limits<long long>::max() / 8,
                           numeric_limits<long long>::max() / 8};
  long long res = tot;
  for (auto r = 0; r <= n; ++r) {
    ctmax(res, tot + pref[r] - ctmin(mval[r & 1], pref[r]));
  }
  cout << res << "\n";
  return 0;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int ttt;
  cin >> ttt;
  while (ttt--) {
    SOLVE();
  }
  return 0;
}
