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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const long long int MOD = 1e9 + 7;
const long long int MOD1 = 998244353;
const long long int N = 2e5 + 5;
const long long int INF = 1000111000111000111LL;
const long double EPS = 1e-12;
const long double PI = 3.141592653589793116;
long long int r1(vector<long long int> v) {
  long long int ans = 0;
  vector<long long int> dp;
  for (long long int i = 0; i + 1 < (long long int)((v).size()); i += 2) {
    dp.push_back(v[i + 1] - v[i]);
  }
  long long int n = (long long int)((dp).size());
  long long int cur = 0;
  for (long long int i = 0; i < n; i++) {
    cur += dp[i];
    if (cur < 0) cur = 0;
    ans = max(ans, cur);
  }
  return ans;
}
long long int r2(vector<long long int> v) {
  long long int ans = 0;
  vector<long long int> dp;
  for (long long int i = 1; i < (long long int)((v).size()); i += 2) {
    dp.push_back(v[i - 1] - v[i]);
  }
  long long int n = (long long int)((dp).size());
  long long int cur = 0;
  for (long long int i = 0; i < n; i++) {
    cur += dp[i];
    if (cur < 0) cur = 0;
    ans = max(ans, cur);
  }
  return ans;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = (0); i < (n); i++) cin >> v[i];
  if (n == 1) {
    cout << v[0] << '\n';
    return;
  }
  vector<long long int> v1 = v;
  reverse((v1).begin(), (v1).end());
  v1.pop_back();
  reverse((v1).begin(), (v1).end());
  long long int sum = 0;
  for (long long int i = 0; i < n; i += 2) {
    sum += v[i];
  }
  cout << sum + max(r2(v1), r1(v)) << '\n';
}
int32_t main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
