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;
vector<long long> a;
long long n;
long long kad(int start) {
  int l = start;
  long long max_so_far = LLONG_MIN;
  long long mx = 0;
  int L, R;
  for (int i = start; i < n; i++) {
    if (i % 2 == 0)
      mx -= a[i];
    else
      mx += a[i];
    if ((i - l + 1) % 2 == 1) continue;
    if (mx > max_so_far) {
      max_so_far = mx;
      L = l, R = i;
    }
    if (mx < 0) {
      l = i + 1;
      mx = 0;
    }
  }
  if (max_so_far <= 0) {
    long long ans = 0;
    for (int i = 0; i < n; i++) ans += a[i] * (i % 2 == 0);
    return ans;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (i >= L && i <= R)
      ans += a[i] * (long long)(i % 2 == 1);
    else
      ans += a[i] * (long long)(i % 2 == 0);
  }
  return ans;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    a = vector<long long>(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    long long ans = kad(0);
    ans = max(ans, kad(1));
    cout << ans << '\n';
  }
}
