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;
using ll = long long;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    ll sum = 0;
    vector<vector<ll>> dp(n);
    vector<ll> v(n);
    for (ll i = 0; i < n; ++i) {
      cin >> v[i];
      sum += v[i] * ((i + 1) % 2);
      dp[i].push_back(0);
      dp[i].push_back(0);
    }
    if (n == 1) {
      cout << sum << endl;
    } else {
      dp[1][1] = max(dp[1][1], -v[0] + v[1]);
      dp[1][0] = max(dp[0][0], dp[1][1]);
      for (int i = 2; i < n; ++i) {
        if (i % 2 == 0) {
          dp[i][1] = max(dp[i - 2][1] - v[i] + v[i - 1], dp[i][1]);
          dp[i][0] = max(dp[i - 1][0], dp[i][1]);
        } else {
          dp[i][1] = max(dp[i - 2][1] + v[i] - v[i - 1], dp[i][1]);
          dp[i][0] = max(dp[i - 1][0], dp[i][1]);
        }
      }
      cout << dp[n - 1][0] + sum << endl;
    }
    dp.clear();
    v.clear();
  }
}
