#include <bits/stdc++.h>
using namespace std;
int t;
int n;
int a[200005];
long long oddpre[100005];
long long evenpre[100005];
long long delta1[100005];
long long delta2[100005];
long long sufmx1[100005];
long long sufmx2[100005];
int main() {
  for (cin >> t; t--;) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) {
      if (i & 1)
        evenpre[(i + 1) / 2] = evenpre[i / 2] + a[i];
      else
        oddpre[i / 2] = oddpre[(i - 1) / 2] + a[i];
    }
    for (int i = 1; i <= n / 2; i++) delta1[i] = oddpre[i] - evenpre[i];
    for (int i = 1; i <= (n - 1) / 2; i++)
      delta2[i] = oddpre[i] - evenpre[i + 1];
    sufmx1[n / 2] = delta1[n / 2];
    for (int i = n / 2 - 1; i >= 0; i--)
      sufmx1[i] = max(delta1[i], sufmx1[i + 1]);
    sufmx2[(n - 1) / 2] = delta2[(n - 1) / 2];
    for (int i = (n - 1) / 2 - 1; i >= 0; i--)
      sufmx2[i] = max(delta2[i], sufmx2[i + 1]);
    long long ans = 0;
    for (int i = 1; i < n; i++) {
      if (i & 1)
        ans = max(ans, evenpre[i / 2] - oddpre[i / 2] + sufmx1[i / 2]);
      else
        ans = max(ans, evenpre[i / 2] - oddpre[i / 2 - 1] + sufmx2[i / 2]);
    }
    cout << ans + evenpre[(n + 1) / 2] << endl;
  }
  return 0;
}

### 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 t;
int n;
int a[200005];
long long oddpre[100005];
long long evenpre[100005];
long long delta1[100005];
long long delta2[100005];
long long sufmx1[100005];
long long sufmx2[100005];
int main() {
  for (cin >> t; t--;) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) {
      if (i & 1)
        evenpre[(i + 1) / 2] = evenpre[i / 2] + a[i];
      else
        oddpre[i / 2] = oddpre[(i - 1) / 2] + a[i];
    }
    for (int i = 1; i <= n / 2; i++) delta1[i] = oddpre[i] - evenpre[i];
    for (int i = 1; i <= (n - 1) / 2; i++)
      delta2[i] = oddpre[i] - evenpre[i + 1];
    sufmx1[n / 2] = delta1[n / 2];
    for (int i = n / 2 - 1; i >= 0; i--)
      sufmx1[i] = max(delta1[i], sufmx1[i + 1]);
    sufmx2[(n - 1) / 2] = delta2[(n - 1) / 2];
    for (int i = (n - 1) / 2 - 1; i >= 0; i--)
      sufmx2[i] = max(delta2[i], sufmx2[i + 1]);
    long long ans = 0;
    for (int i = 1; i < n; i++) {
      if (i & 1)
        ans = max(ans, evenpre[i / 2] - oddpre[i / 2] + sufmx1[i / 2]);
      else
        ans = max(ans, evenpre[i / 2] - oddpre[i / 2 - 1] + sufmx2[i / 2]);
    }
    cout << ans + evenpre[(n + 1) / 2] << endl;
  }
  return 0;
}

```