#include <bits/stdc++.h>
using namespace std;
namespace IO {
template <typename T>
inline void read(T &x) {
  x = 0;
  long long f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x = x * f;
}
}  // namespace IO
using namespace IO;
const long long maxn = 200000;
long long dp[maxn + 10][4];
long long a[maxn + 10];
void DP(long long &x, long long y) { x = max(x, y); }
signed main() {
  long long _;
  read(_);
  while (_--) {
    long long n;
    read(n);
    for (long long i = 0; i <= n; i++) {
      dp[i][0] = dp[i][1] = dp[i][2] = 0;
      a[i] = 0;
    }
    for (long long i = 0; i < n; i++) {
      read(a[i]);
    }
    for (long long i = 0; i < n; i++) {
      if (i % 2 == 1) {
        DP(dp[i + 1][0], dp[i][0]);
        if (i + 2 <= n) DP(dp[i + 2][1], max(dp[i][1] + a[i], dp[i][0] + a[i]));
        DP(dp[i + 1][2], max(dp[i][1], max(dp[i][0], dp[i][2])));
      } else {
        DP(dp[i + 1][0], dp[i][0] + a[i]);
        if (i + 2 <= n) DP(dp[i + 2][1], max(dp[i][1], dp[i][0]) + a[i + 1]);
        DP(dp[i + 1][2], max(dp[i][1], max(dp[i][0], dp[i][2])) + a[i]);
      }
    }
    cout << max(dp[n][1], max(dp[n][2], dp[n][0])) << endl;
  }
}

### Prompt

Please create a solution in Cpp to 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;
namespace IO {
template <typename T>
inline void read(T &x) {
  x = 0;
  long long f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x = x * f;
}
}  // namespace IO
using namespace IO;
const long long maxn = 200000;
long long dp[maxn + 10][4];
long long a[maxn + 10];
void DP(long long &x, long long y) { x = max(x, y); }
signed main() {
  long long _;
  read(_);
  while (_--) {
    long long n;
    read(n);
    for (long long i = 0; i <= n; i++) {
      dp[i][0] = dp[i][1] = dp[i][2] = 0;
      a[i] = 0;
    }
    for (long long i = 0; i < n; i++) {
      read(a[i]);
    }
    for (long long i = 0; i < n; i++) {
      if (i % 2 == 1) {
        DP(dp[i + 1][0], dp[i][0]);
        if (i + 2 <= n) DP(dp[i + 2][1], max(dp[i][1] + a[i], dp[i][0] + a[i]));
        DP(dp[i + 1][2], max(dp[i][1], max(dp[i][0], dp[i][2])));
      } else {
        DP(dp[i + 1][0], dp[i][0] + a[i]);
        if (i + 2 <= n) DP(dp[i + 2][1], max(dp[i][1], dp[i][0]) + a[i + 1]);
        DP(dp[i + 1][2], max(dp[i][1], max(dp[i][0], dp[i][2])) + a[i]);
      }
    }
    cout << max(dp[n][1], max(dp[n][2], dp[n][0])) << endl;
  }
}

```