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;
const long long INF = 2e18;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    long long tab[n], s1 = 0, s2 = 0;
    vector<long long> v1, v2;
    for (i = 0; i < n; i++) {
      cin >> tab[i];
      if (i % 2 == 0) s1 += tab[i];
    }
    if (n == 1) {
      cout << tab[0] << "\n";
      continue;
    }
    for (i = 1; i < n; i++) {
      if (i % 2) {
        v1.push_back(tab[i] - tab[i - 1]);
      } else {
        v2.push_back(tab[i - 1] - tab[i]);
      }
    }
    long long m1 = 0, m2 = 0, k1 = 0, k2 = 0;
    for (i = 0; i < v1.size(); i++) {
      if (i > 0) v1[i] += v1[i - 1];
      m1 = min(v1[i], m1);
      k1 = max(v1[i] - m1, k1);
    }
    for (i = 0; i < v2.size(); i++) {
      if (i > 0) v2[i] += v2[i - 1];
      m2 = min(v2[i], m2);
      k2 = max(v2[i] - m2, k2);
    }
    cout << s1 + max(k1, k2) << "\n";
  }
  return 0;
}
