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::sync_with_stdio(0);
  cin.tie(0);
  int a1;
  cin >> a1;
  for (int f = 0; f < a1; f++) {
    int n;
    cin >> n;
    long long int odd[n / 2 + 3], even[n / 2 + 3], o = 0, e = 0;
    long long int ans = 0;
    for (int i = 0, val; i < n; i++) {
      cin >> val;
      if (i % 2) {
        odd[o++] = val;
      } else {
        even[e++] = val;
        ans += val;
      }
    }
    long long int maone = 0, matwo = 0, val = 0;
    for (int i = 0; i < n / 2; i++) {
      val = max(odd[i] - even[i], val + odd[i] - even[i]);
      maone = max(maone, val);
    }
    val = 0;
    for (int i = 1; i < n / 2; i++) {
      val = max(odd[i - 1] - even[i], val + odd[i - 1] - even[i]);
      matwo = max(matwo, val);
    }
    if (n % 2 && n > 1) {
      matwo = max(
          matwo, max(odd[o - 1] - even[e - 1], val + odd[o - 1] - even[e - 1]));
    }
    cout << ans + max(maone, matwo) << "\n";
  }
  return 0;
}
