#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1000000007;
long long int MX = 40000000000000;
long long int T;
long long int n;
long long int a[200200];
long long int sum[200200];
long long int dp[200200];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i <= n; i++) dp[i] = 0;
    long long int res = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        res += a[i];
        a[i] *= -1;
      }
      if (i == 0)
        sum[i] = a[i];
      else
        sum[i] = sum[i - 1] + a[i];
    }
    long long int mi0 = sum[0];
    long long int mi1 = 0;
    long long int temp = 0;
    for (int i = 1; i < n; i++) {
      if (i % 2 == 1) {
        dp[i] = sum[i] - mi1;
        mi1 = min(mi1, sum[i]);
      } else {
        dp[i] = sum[i] - mi0;
        mi0 = min(mi0, sum[i]);
      }
      dp[i] = max(dp[i], (long long int)0);
      temp = max(temp, dp[i]);
    }
    cout << temp + res << 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;
long long int MOD = 1000000007;
long long int MX = 40000000000000;
long long int T;
long long int n;
long long int a[200200];
long long int sum[200200];
long long int dp[200200];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i <= n; i++) dp[i] = 0;
    long long int res = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        res += a[i];
        a[i] *= -1;
      }
      if (i == 0)
        sum[i] = a[i];
      else
        sum[i] = sum[i - 1] + a[i];
    }
    long long int mi0 = sum[0];
    long long int mi1 = 0;
    long long int temp = 0;
    for (int i = 1; i < n; i++) {
      if (i % 2 == 1) {
        dp[i] = sum[i] - mi1;
        mi1 = min(mi1, sum[i]);
      } else {
        dp[i] = sum[i] - mi0;
        mi0 = min(mi0, sum[i]);
      }
      dp[i] = max(dp[i], (long long int)0);
      temp = max(temp, dp[i]);
    }
    cout << temp + res << endl;
  }
  return 0;
}

```