#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5;
long long a[N], b[N], dp[N], pref[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long ans = 0, sum = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      if (i % 2 == 0) ans += a[i];
    }
    int ind = 0;
    sum = ans;
    for (int i = 0; i < n; i += 2) {
      if (i + 1 < n)
        b[ind] = a[i + 1] - a[i];
      else
        b[ind] = -a[i];
      if (ind)
        pref[ind] = pref[ind - 1] + b[ind];
      else
        pref[ind] = b[ind];
      ind++;
    }
    int m = ind;
    dp[0] = b[0];
    if (b[0] < 0)
      ind = 0;
    else
      ind = -1;
    long long mn = min(0LL, b[0]);
    ans = max(ans, sum + dp[0]);
    for (int i = 1; i < m; ++i) {
      if (ind == -1)
        dp[i] = pref[i];
      else
        dp[i] = pref[i] - pref[ind];
      if (pref[i] < mn) {
        mn = pref[i];
        ind = i;
      }
      ans = max(ans, sum + dp[i]);
    }
    ind = 0;
    for (int i = 1; i + 1 < n; i += 2) {
      b[ind] = a[i] - a[i + 1];
      if (ind)
        pref[ind] = pref[ind - 1] + b[ind];
      else
        pref[ind] = b[ind];
      ind++;
    }
    m = ind;
    dp[0] = b[0];
    if (b[0] < 0)
      ind = 0;
    else
      ind = -1;
    mn = min(0LL, b[0]);
    ans = max(ans, sum + dp[0]);
    for (int i = 1; i < m; ++i) {
      if (ind == -1)
        dp[i] = pref[i];
      else
        dp[i] = pref[i] - pref[ind];
      if (pref[i] < mn) {
        mn = pref[i];
        ind = i;
      }
      ans = max(ans, sum + dp[i]);
    }
    cout << ans << "\n";
  }
}

### 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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5;
long long a[N], b[N], dp[N], pref[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long ans = 0, sum = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      if (i % 2 == 0) ans += a[i];
    }
    int ind = 0;
    sum = ans;
    for (int i = 0; i < n; i += 2) {
      if (i + 1 < n)
        b[ind] = a[i + 1] - a[i];
      else
        b[ind] = -a[i];
      if (ind)
        pref[ind] = pref[ind - 1] + b[ind];
      else
        pref[ind] = b[ind];
      ind++;
    }
    int m = ind;
    dp[0] = b[0];
    if (b[0] < 0)
      ind = 0;
    else
      ind = -1;
    long long mn = min(0LL, b[0]);
    ans = max(ans, sum + dp[0]);
    for (int i = 1; i < m; ++i) {
      if (ind == -1)
        dp[i] = pref[i];
      else
        dp[i] = pref[i] - pref[ind];
      if (pref[i] < mn) {
        mn = pref[i];
        ind = i;
      }
      ans = max(ans, sum + dp[i]);
    }
    ind = 0;
    for (int i = 1; i + 1 < n; i += 2) {
      b[ind] = a[i] - a[i + 1];
      if (ind)
        pref[ind] = pref[ind - 1] + b[ind];
      else
        pref[ind] = b[ind];
      ind++;
    }
    m = ind;
    dp[0] = b[0];
    if (b[0] < 0)
      ind = 0;
    else
      ind = -1;
    mn = min(0LL, b[0]);
    ans = max(ans, sum + dp[0]);
    for (int i = 1; i < m; ++i) {
      if (ind == -1)
        dp[i] = pref[i];
      else
        dp[i] = pref[i] - pref[ind];
      if (pref[i] < mn) {
        mn = pref[i];
        ind = i;
      }
      ans = max(ans, sum + dp[i]);
    }
    cout << ans << "\n";
  }
}

```