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;
long long mod = 1000000007;
long long dx[] = {-1, 1, 0, 0};
long long dy[] = {0, 0, 1, -1};
bool test = 1;
const long long N = 2e5 + 5;
long long n;
long long a[N];
long long cache[N][4];
long long dp(long long idx, long long cur) {
  if (idx >= n) return 0;
  long long &ans = cache[idx][cur];
  if (ans != -1) return ans;
  ans = 0;
  if (cur == 0) {
    ans = max(ans, a[idx] + dp(idx + 2, cur));
    if (idx < n - 2) ans = max(ans, a[idx] + dp(idx + 1, 1));
    ans = max(ans, dp(idx + 1, 2));
  } else if (cur == 1) {
    if (idx < n - 3) ans = max(ans, a[idx] + dp(idx + 2, cur));
    ans = max(ans, a[idx] + dp(idx + 3, 3));
  } else if (cur == 2) {
    ans = max(ans, a[idx] + dp(idx + 2, cur));
    ans = max(ans, a[idx] + dp(idx + 1, 3));
  } else
    ans = max(ans, a[idx] + dp(idx + 2, cur));
  return ans;
}
void solve() {
  cin >> n;
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j < 4; j++) cache[i][j] = -1;
  for (long long i = 0; i < n; i++) cin >> a[i];
  cout << dp(0, 0) << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  if (test) cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
