#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long N;
    cin >> N;
    vector<long long> A(N);
    for (long long i = 0; i < N; i++) cin >> A[i];
    if (N == 1) {
      cout << A[0] << endl;
      continue;
    }
    long long ans = 0;
    vector<long long> B, C;
    for (long long i = 0; i < N; i++) {
      if (i % 2)
        C.push_back(A[i]);
      else {
        B.push_back(A[i]);
        ans += A[i];
      }
    }
    long long mi = 0;
    long long res = -923372036854775807;
    vector<long long> sum, sum1, sum2;
    sum.push_back(0);
    sum1.push_back(0);
    for (long long i = 0; i < C.size(); i++) {
      sum.push_back(C[i] - B[i] + sum.back());
      if (i + 1 < B.size()) sum1.push_back(C[i] - B[i + 1] + sum1.back());
    }
    for (long long i = 0; i < sum.size(); i++) {
      chmin(mi, sum[i]);
      chmax(res, sum[i] - mi);
    }
    mi = 0;
    for (long long i = 0; i < sum1.size(); i++) {
      chmin(mi, sum1[i]);
      chmax(res, sum1[i] - mi);
    }
    cout << ans + res << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long N;
    cin >> N;
    vector<long long> A(N);
    for (long long i = 0; i < N; i++) cin >> A[i];
    if (N == 1) {
      cout << A[0] << endl;
      continue;
    }
    long long ans = 0;
    vector<long long> B, C;
    for (long long i = 0; i < N; i++) {
      if (i % 2)
        C.push_back(A[i]);
      else {
        B.push_back(A[i]);
        ans += A[i];
      }
    }
    long long mi = 0;
    long long res = -923372036854775807;
    vector<long long> sum, sum1, sum2;
    sum.push_back(0);
    sum1.push_back(0);
    for (long long i = 0; i < C.size(); i++) {
      sum.push_back(C[i] - B[i] + sum.back());
      if (i + 1 < B.size()) sum1.push_back(C[i] - B[i + 1] + sum1.back());
    }
    for (long long i = 0; i < sum.size(); i++) {
      chmin(mi, sum[i]);
      chmax(res, sum[i] - mi);
    }
    mi = 0;
    for (long long i = 0; i < sum1.size(); i++) {
      chmin(mi, sum1[i]);
      chmax(res, sum1[i] - mi);
    }
    cout << ans + res << endl;
  }
  return 0;
}

```