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 a[200000];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    if (n == 1) {
      cout << a[0] << '\n';
    } else {
      long long ans = 0;
      for (int i = 0; i < n; i++)
        if (i % 2 == 0) ans += a[i];
      vector<long long> v1(n / 2);
      for (int i = 0; i < n / 2; i++) v1[i] = a[2 * i + 1] - a[2 * i];
      vector<long long> v2((n - 1) / 2);
      for (int i = 0; i < (n - 1) / 2; i++) v2[i] = a[2 * i + 1] - a[2 * i + 2];
      long long mx = v1[0], c_mx = v1[0];
      for (int i = 1; i < n / 2; i++) {
        c_mx = max(c_mx, 0ll) + v1[i];
        mx = max(mx, c_mx);
      }
      if (n > 2) {
        c_mx = v2[0];
        mx = max(mx, c_mx);
        for (int i = 1; i < (n - 1) / 2; i++) {
          c_mx = max(c_mx, 0ll) + v2[i];
          mx = max(mx, c_mx);
        }
      }
      ans += max(mx, 0ll);
      cout << ans << '\n';
    }
  }
}
