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 lint = long long;
const int inf = 1000000007;
const int MOD = 1000000007;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<lint> so(n + 1, 0);
  vector<lint> se(n + 1, 0);
  map<lint, int> odd;
  map<lint, int> even;
  odd[0] = 1;
  even[0] = 1;
  lint add = 0;
  lint res = 0;
  for (int i = 1; i <= n; ++i) {
    if (i % 2 == 0) {
      so[i] = so[i - 1] + a[i - 1];
      se[i] = se[i - 1];
    } else {
      so[i] = so[i - 1];
      se[i] = se[i - 1] + a[i - 1];
    }
    res = max(res, se[i]);
    if (i % 2 == 0)
      odd[so[i] - se[i]] += 1;
    else
      even[so[i] - se[i]] += 1;
    if (i % 2 == 0) {
      auto it = odd.begin();
      add = max(add, so[i] - se[i] - it->first);
    } else {
      auto it = even.begin();
      add = max(add, so[i] - se[i] - it->first);
    }
  }
  cout << res + add << "\n";
}
signed main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) solve();
  return 0;
}
