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;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
long long kadanes(vector<int> &a) {
  int n = (int)a.size();
  long long curr = 0, mx = 0;
  for (int i = 0; i < n; ++i) {
    curr += a[i];
    if (curr < 0) curr = 0;
    mx = max(mx, curr);
  }
  return mx;
}
void solve() {
  long long i, j, n, tot = 0;
  cin >> n;
  vector<int> a(n);
  for (i = 0; i < n; ++i) {
    cin >> a[i];
    if (i % 2 == 0) tot += a[i];
  }
  vector<int> kadane1, kadane2;
  for (i = 0; i < n; i += 2)
    if (i + 1 < n) kadane1.push_back(a[i + 1] - a[i]);
  for (i = 1; i < n; i += 2)
    if (i + 1 < n) kadane2.push_back(a[i] - a[i + 1]);
  cout << tot + max(kadanes(kadane2), kadanes(kadane1)) << "\n";
}
