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;
string char_to_str(char c) {
  string tem(1, c);
  return tem;
}
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  }
  return b;
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  }
  return b;
}
long long max_subarray_sum(vector<long long> v) {
  long long size = v.size();
  long long max_so_far = INT_MIN, max_ending_here = 0;
  for (int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + v[i];
    if (max_so_far < max_ending_here) {
      max_so_far = max_ending_here;
    }
    if (max_ending_here < 0) {
      max_ending_here = 0;
    }
  }
  return max_so_far;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    vector<long long> v;
    long long sum = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      if (i % 2 == 0) {
        sum += a[i];
      }
    }
    long long check = 0;
    if (n % 2 == 0) {
      vector<long long> v1;
      vector<long long> v2;
      for (int i = 0; i < n; i += 2) {
        v1.push_back(a[i + 1] - a[i]);
      }
      for (int i = 1; i < n - 1; i += 2) {
        v2.push_back(a[i] - a[i + 1]);
      }
      check = max(check, max_subarray_sum(v1));
      check = max(check, max_subarray_sum(v2));
    } else {
      vector<long long> v1;
      vector<long long> v2;
      for (int i = 0; i < n - 1; i += 2) {
        v1.push_back(a[i + 1] - a[i]);
      }
      for (int i = 1; i < n; i += 2) {
        v2.push_back(a[i] - a[i + 1]);
      }
      check = max(check, max_subarray_sum(v1));
      check = max(check, max_subarray_sum(v2));
      vector<long long> v3;
      vector<long long> v4;
      reverse(a, a + n);
      for (int i = 0; i < n - 1; i += 2) {
        v3.push_back(a[i + 1] - a[i]);
      }
      for (int i = 1; i < n; i += 2) {
        v4.push_back(a[i] - a[i + 1]);
      }
      check = max(check, max_subarray_sum(v3));
      check = max(check, max_subarray_sum(v4));
    }
    cout << sum + check << endl;
  }
  return 0;
}
