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 <class T>
bool max_self(T& a, const T& b) {
  return b > a ? a = b, true : false;
}
template <class T>
bool min_self(T& a, const T& b) {
  return b < a ? a = b, true : false;
}
const int INF = 1e9 + 7;
const double PI = acos(-1.0);
const double EPS = (1e-9);
long long testcases, N, num[(int)2e5 + 7], opt;
long long get(vector<long long>& a) {
  long long pre = 0, sub = 0;
  for (long long ele : a) {
    pre = max(ele, pre + ele);
    max_self(sub, pre);
  }
  return sub;
}
void solve() {
  vector<long long> even, odd;
  opt = 0;
  for (int idx = 0; idx < N; idx += 2) {
    if (idx + 1 < N) {
      even.push_back(num[idx + 1] - num[idx]);
    }
    opt += num[idx];
  }
  for (int idx = 1; idx < N; idx += 2) {
    if (idx + 1 < N) {
      odd.push_back(num[idx] - num[idx + 1]);
    }
  }
  cout << opt + max(get(even), get(odd)) << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> testcases;
  while (testcases--) {
    cin >> N;
    for (int rep = 0; rep < N; ++rep) cin >> num[rep];
    solve();
  }
  return 0;
}
