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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int tc = 1;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (int i = 0; i < (int)(n); ++i) cin >> arr[i];
    if (n == 1) {
      cout << arr[0] << endl;
      continue;
    }
    if (n == 2) {
      int x = max(arr[0], arr[1]);
      cout << x << endl;
      continue;
    }
    vector<long long> even(n, 0);
    vector<long long> odd(n, 0);
    long long final = 0, ans = 0;
    even[0] = arr[0];
    odd[1] = arr[1];
    for (int i = 2; i < n; i += 2) {
      even[i] += even[i - 2] + arr[i];
      final = even[i];
    }
    for (int i = 3; i < n; i += 2) {
      odd[i] += odd[i - 2] + arr[i];
    }
    ans = final;
    int i = 0, j = 1;
    while (j < n && i < n) {
      long long e = ((j % 2 != 0) ? (j - 1 < 0) ? 0 : even[j - 1] : even[j]) -
                    ((j % 2 != 0)  ? (i - 2 < 0) ? 0 : even[i - 2]
                      : (i - 1 < 0) ? 0
                                   : even[i - 1]);
      long long o = ((j % 2 != 0)  ? odd[j]
                     : (j - 1 < 0) ? 0
                                   : odd[j - 1]) -
                    ((j % 2 != 0)  ? (i - 1 < 0) ? 0 : odd[i - 1]
                      : (i - 2 < 0) ? 0
                                   : even[i - 2]);
      if (o - e <= 0) {
        i = j + 1;
        j = i + 1;
      } else {
        j += 2;
      }
      long long temp = final + (o - e);
      ans = max(ans, temp);
    }
    i = 1, j = 2;
    while (j < n && i < n) {
      long long e = ((j % 2 != 0) ? (j - 1 < 0) ? 0 : even[j - 1] : even[j]) -
                    ((j % 2 != 0)  ? (i - 2 < 0) ? 0 : even[i - 2]
                      : (i - 1 < 0) ? 0
                                   : even[i - 1]);
      long long o = ((j % 2 != 0)  ? odd[j]
                     : (j - 1 < 0) ? 0
                                   : odd[j - 1]) -
                    ((j % 2 != 0)  ? (i - 1 < 0) ? 0 : odd[i - 1]
                      : (i - 2 < 0) ? 0
                                   : odd[i - 2]);
      if (o - e <= 0) {
        i = j + 1;
        j = i + 1;
      } else {
        j += 2;
      }
      long long temp = final + (o - e);
      ans = max(ans, temp);
    }
    cout << ans << endl;
  }
  return 0;
}
