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;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using ull = unsigned long long;
using intt = long long;
using ll = long long;
using ld = long double;
const ll m9 = 998244353;
const ll m7 = 1000000007;
const ll m18 = 1000000000000000000;
ll T, n, a[200001], pr[200001], p[2], sum, ans;
int main() {
  ios_base::sync_with_stdio(0);
  (cin).tie(0);
  (cout).tie(0);
  cin >> T;
  while (T--) {
    sum = ans = 0;
    p[0] = p[1] = 0;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (i & 1) {
        sum += a[i];
        a[i] = -a[i];
      }
    }
    for (int i = 1; i <= n; i++) {
      pr[i] = a[i] + pr[i - 1];
      ans = max(ans, pr[i] - p[i & 1]);
      p[i & 1] = min(p[i & 1], pr[i]);
    }
    cout << sum + ans << '\n';
  }
  return 0;
}
