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;
int t, n;
long long a[200005], s_e[200005], s_o[200005], odd_minus_even[200005], res;
bool isDebug = false;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    res = 0;
    odd_minus_even[0] = 0;
    long long min_odd_minus_even_at_even_left = 0,
              min_odd_minus_even_at_odd_left = 0, best_swap = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (i == 0) {
        s_e[i] = 0;
        s_o[i] = 0;
      } else {
        s_e[i] = s_e[i - 1];
        s_o[i] = s_o[i - 1];
      }
      if (i & 1) {
        s_o[i] += a[i];
      } else {
        s_e[i] += a[i];
      }
      odd_minus_even[i] = s_o[i] - s_e[i];
      if (i & 1) {
        min_odd_minus_even_at_odd_left =
            min(min_odd_minus_even_at_odd_left, odd_minus_even[i]);
        best_swap =
            max(best_swap, odd_minus_even[i] - min_odd_minus_even_at_odd_left);
        if (isDebug) {
          cout << "i=" << i << " is odd : best_swap=" << best_swap
               << " ifSwap=";
          cout << odd_minus_even[i] - min_odd_minus_even_at_odd_left;
          cout << " min_dis_at_odd_left=" << min_odd_minus_even_at_odd_left
               << " dis_at_this=" << odd_minus_even[i] << "\n";
        }
      } else {
        min_odd_minus_even_at_even_left =
            min(min_odd_minus_even_at_even_left, odd_minus_even[i]);
        best_swap =
            max(best_swap, odd_minus_even[i] - min_odd_minus_even_at_even_left);
        if (isDebug) {
          cout << "i=" << i << " is even: best_swap=" << best_swap
               << " ifSwap=";
          cout << odd_minus_even[i] - min_odd_minus_even_at_even_left;
          cout << " min_dis_at_even_left =" << min_odd_minus_even_at_even_left
               << " dis_at_this=" << odd_minus_even[i] << "\n";
        }
      }
    }
    res = s_e[n - 1] + best_swap;
    if (isDebug) {
      cout << "======>>>> FINAL RESULT : " << res << " <<<<<=====\n";
    } else {
      cout << res << "\n";
    }
  }
  return 0;
}
