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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    int N;
    cin >> N;
    vector<long long int> a_vec(N, 0);
    long long int even = 0;
    for (int n = 0; n < N; n++) {
      long long int a;
      cin >> a;
      a_vec[n] = a;
      if (n % 2 == 0) even += a;
    }
    long long int cm = 0;
    long long int bm = 0;
    for (int i = 1; i < N; i = i + 2) {
      long long int x = a_vec[i] - a_vec[i - 1];
      cm = max(0ll, cm + x);
      bm = max(bm, cm);
    }
    cm = 0;
    for (int i = 2; i < N; i = i + 2) {
      long long int x = a_vec[i - 1] - a_vec[i];
      cm = max(0ll, cm + x);
      bm = max(bm, cm);
    }
    long long int res = even + bm;
    cout << res << "\n";
  }
}
