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 = 1e18;
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n);
  vector<vector<long long>> dp(n, vector<long long>(3, -INF));
  long long ans = -INF;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i % 2 == 0) {
      dp[i][0] = a[i];
      if (i > 0) dp[i][0] += dp[i - 1][0];
      if (i) {
        dp[i][1] = a[i - 1];
        if (i - 2 >= 0) dp[i][1] += max(dp[i - 2][0], dp[i - 2][1]);
      }
      dp[i][2] = a[i];
      if (i) dp[i][2] += dp[i - 1][1];
      if (i) {
        dp[i][2] = max(dp[i][2], dp[i - 1][2] + a[i]);
      }
    } else {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = a[i];
      if (i - 2 >= 0) {
        dp[i][1] += max(dp[i - 2][0], dp[i - 2][1]);
      }
      dp[i][2] = dp[i - 1][2];
      dp[i][2] = max(dp[i][2], dp[i - 1][1]);
    }
    ans = max(dp[i][0], ans);
    ans = max(ans, dp[i][1]);
    ans = max(ans, dp[i][2]);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
