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;
const int MX = 500005;
const long long MOD = 1000000007;
void setIO(string name) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
int t;
int n;
void solve() {
  cin >> n;
  long long x = 0, even = 0, odd = 0;
  long long sum = 0, improv = 0;
  for (int i = 0; i < (n); i++) {
    int a;
    cin >> a;
    if (i % 2 == 0) {
      x -= a;
      improv = max(improv, x - even);
      even = min(x, even);
      sum += a;
    }
    if (i % 2 == 1) {
      x += a;
      improv = max(improv, x - odd);
      odd = min(x, odd);
    }
  }
  cout << sum + improv << endl;
}
int main() {
  cin >> t;
  for (int i = 0; i < (t); i++) {
    solve();
  }
}
