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;
const long double pi = 3.14159265358979323846;
long long MOD = 1e9 + 7;
const char nl = '\n';
long long gcd(long long a, long long b) {
  if (a < b) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long sq(long long a) {
  long long ans = (1ll * a * a);
  return ans;
}
vector<int> Divisors(int n) {
  vector<int> a;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        a.push_back(i);
      } else {
        a.push_back(i);
        a.push_back(n / i);
      }
    }
  }
  sort(a.begin(), a.end());
  return a;
}
void solve() {
  long long n;
  cin >> n;
  int i;
  long long a[n];
  long long even = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 0) {
      even += a[i];
    }
  }
  long long f = 0;
  long long m = 0;
  for (i = 0; i < n - 1; i += 2) {
    m += a[i + 1] - a[i];
    f = max(f, m);
    if (m < 0) {
      m = 0;
    }
  }
  long long ff = 0;
  long long mm = 0;
  for (i = 1; i < n - 1; i += 2) {
    mm += a[i] - a[i + 1];
    ff = max(ff, mm);
    if (mm < 0) {
      mm = 0;
    }
  }
  cout << even + max(f, ff) << nl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long TC = 1;
  cin >> TC;
  while (TC--) {
    solve();
  }
}
