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 <class T>
using VV = vector<vector<T>>;
template <class T>
inline T SMIN(T& a, const T b) {
  return a = (a > b) ? b : a;
}
template <class T>
inline T SMAX(T& a, const T b) {
  return a = (a < b) ? b : a;
}
template <class Iter>
void __kumaerrc(Iter begin, Iter end) {
  for (; begin != end; ++begin) {
    cerr << *begin << ',';
  }
  cerr << endl;
}
void __kumaerr(istream_iterator<string> it) {
  (void)it;
  cerr << endl;
}
template <typename T, typename... Args>
void __kumaerr(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << "=" << a << ", ", __kumaerr(++it, args...);
}
template <typename S, typename T>
std::ostream& operator<<(std::ostream& _os, const std::pair<S, T>& _p) {
  return _os << "{" << _p.first << ',' << _p.second << "}";
}
const int MAX_N = 1e6 + 1;
long long A[MAX_N];
int N;
long long f(vector<long long> bs) {
  int N = int((bs).size());
  long long cur = 0, mn = 0, res = 0;
  for (int i = 0; i < (N); ++i) {
    cur += bs[i];
    SMAX(res, cur - mn);
    SMIN(mn, cur);
  }
  return res;
}
void solve() {
  long long base = 0;
  for (int i = 0; i < (N); ++i)
    if (i % 2 == 0) base += A[i];
  vector<long long> bs;
  for (int i = 0; i < (N / 2); ++i) bs.push_back(A[2 * i + 1] - A[2 * i]);
  long long res = base;
  SMAX(res, base + f(bs));
  bs.clear();
  for (int i = (1); i <= ((N - 1) / 2); ++i)
    bs.push_back(A[2 * i - 1] - A[2 * i]);
  SMAX(res, base + f(bs));
  cout << res << endl;
}
const long long Inf = 1e18;
long long dp[MAX_N][4];
void solve_dp() {
  for (int i = 0; i <= (N); ++i)
    for (int j = 0; j < (4); ++j) dp[i][j] = -Inf;
  dp[0][0] = dp[0][2] = dp[0][3] = 0;
  for (int i = 0; i < (N); ++i)
    for (int j = 0; j < (4); ++j) {
      if (j == 0) {
        if (i % 2 == 0) {
          SMAX(dp[i + 1][j], dp[i][j] + A[i]);
          SMAX(dp[i + 1][1], dp[i][j] + A[i]);
        } else {
          SMAX(dp[i + 1][j], dp[i][j]);
          SMAX(dp[i + 1][2], dp[i][j]);
        }
      }
      if (j == 1 || j == 2) {
        if (i % 2 == 0) {
          SMAX(dp[i + 1][j], dp[i][j]);
          if (j == 1) SMAX(dp[i + 1][3], dp[i][j]);
        } else {
          SMAX(dp[i + 1][j], dp[i][j] + A[i]);
          if (j == 2) SMAX(dp[i + 1][3], dp[i][j] + A[i]);
        }
      }
      if (j == 3) {
        if (i % 2 == 0) {
          SMAX(dp[i + 1][j], dp[i][j] + A[i]);
        } else {
          SMAX(dp[i + 1][j], dp[i][j]);
        }
      }
    }
  long long res = 0;
  if (N % 2 == 0) dp[N][1] = -Inf;
  for (int j = 0; j < (4); ++j) {
    SMAX(res, dp[N][j]);
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << setprecision(12) << fixed;
  int t;
  cin >> t;
  while (t--) {
    cin >> N;
    for (int i = 0; i < (N); ++i) cin >> A[i];
    solve();
  }
  return 0;
}
