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;
#pragma optimise("-O3")
template <int D, typename T>
struct Vec : public vector<Vec<D - 1, T>> {
  static_assert(D >= 1, "Vector dimension must be greater than zero!");
  template <typename... Args>
  Vec(int n = 0, Args... args)
      : vector<Vec<D - 1, T>>(n, Vec<D - 1, T>(args...)) {}
};
template <typename T>
struct Vec<1, T> : public vector<T> {
  Vec(int n = 0, const T& val = T()) : vector<T>(n, val) {}
};
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (y == 0) return 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modinv(long long int a, long long int p) {
  return power(a, p - 2, p);
}
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1},
          dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
template <class T>
bool ckmin(T& a, T& b) {
  bool B = a > b;
  a = min(a, b);
  return B;
}
template <class T>
bool ckmax(T& a, T& b) {
  bool B = a < b;
  a = max(a, b);
  return B;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n + 1);
  for (long long int i = 1; i <= n; i += 1) {
    cin >> a[i];
  }
  Vec<2, long long int> dp(n + 1, 4, 0);
  dp[0][0] = 0;
  dp[0][1] = 0;
  dp[0][2] = 0;
  dp[0][3] = 0;
  for (long long int i = 1; i <= n; i += 1) {
    if (i % 2) {
      dp[i][0] = dp[i - 1][0] + a[i];
      dp[i][1] = dp[i - 1][2];
      dp[i][2] = dp[i - 1][1];
      dp[i][3] = dp[i - 1][3] + a[i];
      dp[i][1] = max(dp[i][1], dp[i - 1][0]);
      dp[i][3] = max(dp[i - 1][2] + a[i], dp[i][3]);
    } else {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = dp[i - 1][2] + a[i];
      dp[i][2] = dp[i - 1][1] + a[i];
      dp[i][3] = dp[i - 1][3];
      dp[i][1] = max(dp[i][1], dp[i - 1][0] + a[i]);
      dp[i][3] = max(dp[i - 1][2], dp[i][3]);
    }
  }
  cout << max({dp[n][0], dp[n][2], dp[n][3]}) << '\n';
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T = 1;
  cin >> T;
  for (long long int c = 0; c < T; c++) {
    solve();
  }
  return 0;
}
