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>
istream& operator>>(istream& is, vector<T>& a) {
  for (T& ai : a) is >> ai;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> const& a) {
  os << "[ ";
  for (const T& ai : a) os << ai << " ";
  return os << "]";
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& a) {
  return is >> a.first >> a.second;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> const& a) {
  return os << "(" << a.first << ", " << a.second << ")";
}
long long f() {
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  long long r = 0, s = 0;
  vector<long long> lo(2);
  for (int i = 0; i < n; ++i) {
    int p = i & 1;
    s += (2 * p - 1) * a[i];
    lo[p] = min(lo[p], s);
    r = max(r, s - lo[p]);
  }
  for (int i = 0; i < n; i += 2) r += a[i];
  return r;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) cout << f() << "\n";
  return 0;
}
