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 mod = 1e9 + 7;
const long double PI = 3.14159265358979;
long long binpow(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool parity(long long x, long long y) {
  bool f = ((x ^ y) < 0);
  return !f;
}
long long ans, n;
vector<long long> v;
long long kadane(vector<long long> test) {
  int sz = test.size();
  long long ans = 0, curr = 0, last = 0;
  for (int i = 0; i < sz; i++) {
    curr += test[i];
    ans = max(ans, curr);
    if (curr < 0) curr = 0;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    long long i, j;
    v.assign(n, 0);
    for (i = 0; i <= n - 1; i++) cin >> v[i];
    long long sum = 0;
    vector<long long> odd, even;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        even.emplace_back(v[i]);
        sum += v[i];
      } else
        odd.emplace_back(v[i]);
    }
    vector<long long> diff;
    if (n % 2 == 0) {
      for (int i = 0; i < n / 2; i++) diff.emplace_back(odd[i] - even[i]);
      ans = kadane(diff);
      diff.clear();
      for (int i = 0; i < n / 2 - 1; i++)
        diff.emplace_back(odd[i] - even[i + 1]);
      ans = max(ans, kadane(diff));
      cout << sum + ans << '\n';
    } else {
      for (int i = 0; i < n / 2; i++) diff.emplace_back(odd[i] - even[i]);
      ans = kadane(diff);
      diff.clear();
      for (int i = 0; i < n / 2; i++) diff.emplace_back(odd[i] - even[i + 1]);
      ans = max(ans, kadane(diff));
      cout << sum + ans << '\n';
    }
  }
  return 0;
}
