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;
using ull = unsigned long long;
using ll = long long;
using ld = long double;
string to_string(char s) {
  string res = "'''";
  res[1] = s;
  return res;
}
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = 1;
  string res = "{";
  for (const auto &x : v) {
    if (!first) res += ", ";
    first = 0;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug() { cout << "\n"; }
template <typename Head, typename... Tail>
void debug(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug(T...);
}
const long long int M = 1e9 + 7;
inline long long int add(long long int a, long long int b,
                         long long int p = M) {
  a = a + b;
  if (a >= p) a -= p;
  return a;
}
inline long long int sub(long long int a, long long int b,
                         long long int p = M) {
  a = a - b;
  if (a < 0) a += p;
  return a;
}
inline long long int mul(long long int a, long long int b,
                         long long int p = M) {
  a = (a * 1ll * b) % p;
  return a;
}
const ld pi = acos(-1), eps = 1e-9;
const ll inf = 1e18;
const long long int N = 1;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (auto &c : v) cin >> c;
  if (n == 1) {
    cout << v[0];
    return;
  }
  vector<long long int> w;
  long long int sum = 0;
  for (long long int i = 0; i < n; ++i) {
    if (i % 2 == 0)
      sum += v[i];
    else
      w.push_back(v[i] - v[i - 1]);
  }
  auto get = [&w]() {
    long long int best = 0, sum = 0;
    for (auto c : w) {
      sum = max(sum + c, c);
      best = max(best, sum);
    }
    return best;
  };
  long long int ans = sum + get();
  w.clear();
  for (long long int i = 1; i + 1 < n; i += 2) {
    w.push_back(v[i] - v[i + 1]);
  }
  ans = max(ans, sum + get());
  cout << ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  cin >> t;
  for (long long int z = 1; z <= t; ++z) {
    solve();
    cout << "\n";
  }
  return 0;
}
