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 ll = long long;
using ld = long double;
using namespace std::chrono;
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long first = 0;
  cerr << '{';
  for (auto &i : x) cerr << (first++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
ll power(ll a, ll p, ll m = 1000000007) {
  ll res = 1;
  while (p > 0) {
    if (p & 1) res = (res * a) % m;
    a = (a * a) % m;
    p >>= 1;
  }
  return res;
}
long long ceil(long long a, long long b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
long long pow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = pow(a, b / 2);
  res = res * res;
  if (b & 1) res *= a;
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    for (long long i = 0; i < n; i++) cin >> v[i];
    long long total = 0;
    vector<long long> l, r;
    for (long long i = 0; i < n; i++) {
      if (i & 1 ^ 1) total += v[i];
    }
    for (long long i = 0; i + 1 < n; i += 2) {
      l.push_back(v[i + 1] - v[i]);
    }
    for (long long i = 1; i + 1 < n; i += 2) {
      r.push_back(v[i] - v[i + 1]);
    }
    long long left = 0;
    long long right = 0;
    long long ans1 = 0, ans2 = 0;
    for (auto a : l) {
      left += a;
      left = max(left, 0ll);
      ans1 = max(ans1, left);
    }
    for (auto a : r) {
      right += a;
      right = max(right, 0ll);
      ans2 = max(ans2, right);
    }
    cout << total + max(ans1, ans2) << '\n';
  }
  return 0;
}
