#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int a[n], ar[n];
    for (int i = 0; i < n; i += 1) cin >> ar[i];
    if (n == 1) {
      cout << ar[0] << endl;
      continue;
    }
    if (n == 2) {
      cout << max(ar[0], ar[1]) << endl;
      continue;
    }
    vector<int> v0, v1;
    for (int i = 1; i < n; i += 2) v0.push_back(ar[i] - ar[i - 1]);
    for (int i = 2; i < n; i += 2) v1.push_back(ar[i - 1] - ar[i]);
    long long int tot0 = -1 * 1000000007, tot1 = -1 * 1000000007;
    long long int me = 0;
    for (int i = 0; i < v0.size(); i += 1) {
      me += v0[i];
      if (tot0 < me) {
        tot0 = me;
      }
      if (me < 0) {
        me = 0;
      }
    }
    me = 0;
    for (int i = 0; i < v1.size(); i += 1) {
      me += v1[i];
      if (tot1 < me) {
        tot1 = me;
      }
      if (me < 0) {
        me = 0;
      }
    }
    if (tot0 < 0) tot0 = 0;
    if (tot1 < 0) tot1 = 0;
    long long int ans = 0;
    for (int i = 0; i < n; i += 2) ans += ar[i];
    cout << ans + max(tot0, tot1) << endl;
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int a[n], ar[n];
    for (int i = 0; i < n; i += 1) cin >> ar[i];
    if (n == 1) {
      cout << ar[0] << endl;
      continue;
    }
    if (n == 2) {
      cout << max(ar[0], ar[1]) << endl;
      continue;
    }
    vector<int> v0, v1;
    for (int i = 1; i < n; i += 2) v0.push_back(ar[i] - ar[i - 1]);
    for (int i = 2; i < n; i += 2) v1.push_back(ar[i - 1] - ar[i]);
    long long int tot0 = -1 * 1000000007, tot1 = -1 * 1000000007;
    long long int me = 0;
    for (int i = 0; i < v0.size(); i += 1) {
      me += v0[i];
      if (tot0 < me) {
        tot0 = me;
      }
      if (me < 0) {
        me = 0;
      }
    }
    me = 0;
    for (int i = 0; i < v1.size(); i += 1) {
      me += v1[i];
      if (tot1 < me) {
        tot1 = me;
      }
      if (me < 0) {
        me = 0;
      }
    }
    if (tot0 < 0) tot0 = 0;
    if (tot1 < 0) tot1 = 0;
    long long int ans = 0;
    for (int i = 0; i < n; i += 2) ans += ar[i];
    cout << ans + max(tot0, tot1) << endl;
  }
}

```