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 int MAX3(long long int x, long long int y, long long int z) {
  long long int t = max(x, y);
  return max(t, z);
}
long long int power(long long int x, long long int n) {
  long long int i, ans = 1;
  for (i = 1; i <= (n); i += 1) {
    ans = ans * x;
  }
  return ans;
}
void solve() {
  long long int n, i, sum = 0;
  vector<long long int> a;
  cin >> n;
  for (i = 0; i < (n); i += 1) {
    long long int x;
    cin >> x;
    a.push_back(x);
    if (i % 2 == 0) sum += x;
  }
  vector<long long int> a1, a2;
  long long int m1 = 0, m2 = 0;
  for (i = 0; i < (n - 1); i += 1) {
    long long int t = -a[i] + a[i + 1];
    i++;
    a1.push_back(t);
  }
  for (i = 1; i < (n - 1); i += 1) {
    long long int t = -a[i + 1] + a[i];
    i++;
    a2.push_back(t);
  }
  long long int s = a1.size();
  long long int max_so_far = 0;
  long long int max_ending_here = 0;
  for (i = 0; i < (s); i += 1) {
    max_ending_here += a1[i];
    max_so_far = max(max_so_far, max_ending_here);
    if (max_ending_here < 0) max_ending_here = 0;
  }
  m1 = max_so_far;
  max_so_far = 0;
  max_ending_here = 0;
  s = a2.size();
  for (i = 0; i < (s); i += 1) {
    max_ending_here += a2[i];
    max_so_far = max(max_so_far, max_ending_here);
    if (max_ending_here < 0) max_ending_here = 0;
  }
  m2 = max_so_far;
  long long int ans = max(m1, m2);
  cout << ans + sum << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
