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 N = 2e5 + 3, mod = 1e9 + 7, inf = 3e18 + 7;
long long tc, n, m, a, b, c, d, i, j, k;
long long A[N], odd[N], even[N];
string s[N];
vector<long long> vec;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n;
    long long sum = 0, ans = 0, mx;
    for (i = 1; i <= n; i++) cin >> A[i], sum += (i % 2 ? A[i] : 0);
    A[n + 1] = A[n];
    for (i = 2; i <= n; i += 2) even[i] = even[i - 2] + A[i] - A[i - 1];
    for (i = 2; i <= n; i += 2) odd[i] = odd[i - 2] + A[i] - A[i + 1];
    if (n % 2) n--;
    for (i = n, mx = even[n]; i > 0; i -= 2)
      mx = max(mx, even[i]), ans = max(ans, mx - even[i - 2]);
    for (i = n, mx = odd[n]; i > 0; i -= 2)
      mx = max(mx, odd[i]), ans = max(ans, mx - odd[i - 2]);
    cout << ans + sum << endl;
  }
}
