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 int inf = 0x3f3f3f3f3f3f3f3fll;
long long int maxSubArraySum(vector<long long int> a, long long int size) {
  long long int max_so_far = -1, max_ending_here = 0;
  for (long long int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
int main() {
  int tt;
  cin >> tt;
  for (int ttt = 1; ttt <= tt; ttt++) {
    long long int n;
    cin >> n;
    long long int arr[n];
    long long int res = 0;
    for (long long int i = 0; i < n; i++) cin >> arr[i];
    for (long long int i = 0; i < n; i += 2) {
      res += arr[i];
    }
    vector<long long int> v1, v2;
    long long int ans1 = 0, ans2 = 0;
    for (long long int i = 0; i < n - 1; i += 2)
      v1.push_back(arr[i + 1] - arr[i]);
    ans1 = maxSubArraySum(v1, v1.size());
    for (long long int i = 1; i < n - 1; i += 2)
      v2.push_back(arr[i] - arr[i + 1]);
    ans2 = maxSubArraySum(v2, v2.size());
    long long int ans3 = 0;
    cout << res + max(ans1, max(ans3, ans2)) << endl;
  }
  return 0;
}
