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;
template <typename T, size_t N>
long long SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
long long SIZE(const T &t) {
  return t.size();
}
string to_string(const string s, long long x1 = 0, long long x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template <size_t N>
string to_string(const bitset<N> &b, long long x1 = 0, long long x2 = 1e9) {
  string t = "";
  for (long long __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(const A(&v), long long x1 = 0, long long x2 = 1e9,
                 C... coords);
long long l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(const pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(const A(&v), long long x1, long long x2, C... coords) {
  long long rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgm() { ; }
template <typename Heads, typename... Tails>
void dbgm(Heads H, Tails... T) {
  cout << to_string(H) << " | ";
  dbgm(T...);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, ans = 0, sum = 0, mx1 = 0, mx2 = 0;
    cin >> n;
    vector<long long> a(n), od, ev;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) ans += a[i];
    }
    vector<long long> p, q;
    for (long long i = 0; i + 1 < n; i += 2) {
      p.push_back(a[i + 1] - a[i]);
    }
    for (long long i = 1; i + 1 < n; i += 2) {
      q.push_back(a[i] - a[i + 1]);
    }
    if (p.size() > 0) {
      mx1 = p[0];
      sum = 0;
      for (long long i = 0; i < p.size(); i++) {
        sum += p[i];
        mx1 = max(mx1, sum);
        sum = max(sum, 0ll);
      }
    }
    if (q.size() > 0) {
      mx2 = q[0];
      sum = 0;
      for (long long i = 0; i < q.size(); i++) {
        sum += q[i];
        mx2 = max(mx2, sum);
        sum = max(sum, 0ll);
      }
    }
    ans = max(ans, ans + max(mx1, mx2));
    cout << ans << '\n';
  }
  return 0;
}
