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 arr[100005];
vector<int> prime;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, i, j, k, l;
    long long int ans = 0, cc = 0, MX = 0;
    cin >> n;
    vector<long long int> vec, vec2;
    long long int arr[n];
    for (i = 0; i < n; i++) {
      cin >> arr[i];
      if (i % 2)
        vec.push_back(arr[i] - arr[i - 1]);
      else {
        ans += arr[i];
        if (i) vec2.push_back(arr[i - 1] - arr[i]);
      }
    }
    for (i = 0; i < vec.size(); i++)
      cc += vec[i], cc = max(0ll, cc), MX = max(MX, cc);
    cc = 0;
    for (i = 0; i < vec2.size(); i++)
      cc += vec2[i], cc = max(0ll, cc), MX = max(MX, cc);
    cout << (ans + MX) << endl;
  }
  return 0;
}
