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;
long long solveOdd(vector<long long> &a) {
  long long n = a.size();
  vector<long long> pref_raz(n + 1), suf_max(n + 10, -2e18);
  for (long long i = 1; i < n; i += 2) {
    if (i == 1)
      pref_raz[i] = a[i] - a[i - 1];
    else
      pref_raz[i] = pref_raz[i - 2] + a[i] - a[i - 1];
  }
  long long start = n - 1;
  if (start % 2 == 0) --start;
  for (long long i = start; i >= 0; i -= 2) {
    suf_max[i] = max(suf_max[i + 2], pref_raz[i]);
  }
  long long mx = -2e18;
  for (long long i = 0; i < n; i += 2) {
    if (i == n - 1) continue;
    if (i == 0) {
      mx = max(mx, suf_max[1]);
    } else {
      mx = max(mx, suf_max[i + 1] - pref_raz[i - 1]);
    }
  }
  return mx;
}
long long solveEven(vector<long long> &a) {
  long long n = a.size();
  vector<long long> pref_raz(n + 1), suf_max(n + 10, -2e18);
  for (long long i = 0; i < n; i += 2) {
    if (i == 0)
      pref_raz[i] = 0;
    else
      pref_raz[i] = pref_raz[i - 2] + a[i - 1] - a[i];
  }
  long long start = n - 1;
  if (start % 2 == 1) --start;
  for (long long i = start; i >= 0; i -= 2) {
    suf_max[i] = max(suf_max[i + 2], pref_raz[i]);
  }
  long long mx = -2e18;
  for (long long i = 1; i < n; i += 2) {
    if (i == n - 1) continue;
    if (i == 0) {
      mx = max(mx, suf_max[1]);
    } else {
      mx = max(mx, suf_max[i + 1] - pref_raz[i - 1]);
    }
  }
  return mx;
}
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    long long sum = 0;
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
      if (i % 2 == 0) sum += a[i];
    }
    long long mx = solveOdd(a);
    mx = max(mx, solveEven(a));
    cout << sum + max(0LL, mx) << '\n';
  }
}
