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 int MAXN = 310000;
const int MAXK = 18;
const long long INF = 1e15;
const long long MOD = 998244353;
vector<int> V[MAXN];
long long N, a, b, c, d, T, Q, M, x, y, z, u, v, w;
long long A[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) {
    cin >> N;
    a = 0;
    for (int i = 1; i <= N; ++i) {
      cin >> A[i];
      if ((i % 2)) {
        a += A[i];
        A[i] = -A[i];
      }
    }
    b = 0;
    long long largodd = -INF;
    long long largeven = -INF;
    for (int i = 1; i <= N; ++i) A[i] += A[i - 1];
    for (int i = N; i >= 0; --i) {
      if (i % 2 == 1) {
        b = max(b, largodd - A[i]);
        largodd = max(largodd, A[i]);
      } else {
        b = max(b, largeven - A[i]);
        largeven = max(largeven, A[i]);
      }
    }
    cout << a + b << '\n';
  }
}
