#include <bits/stdc++.h>
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
using namespace std;
const long long maxn = 1e3 + 7;
const double eps = 1e-10;
const long long mod = 1e9 + 7;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> num(n);
    long long ans = 0;
    for (long long i = 0; i < (long long)(n); i++) {
      cin >> num[i];
      if ((i & 1) == 0) ans += num[i];
    }
    long long sum = 0;
    long long cas = 0;
    vector<long long> dp;
    long long i = 0;
    while (i + 1 < n) {
      dp.push_back(num[i + 1] - num[i]);
      i += 2;
    }
    for (auto &x : dp) {
      sum += x;
      cas = max(sum, cas);
      if (sum < 0) sum = 0;
    }
    dp.clear();
    i = n - 1;
    if (i & 1) i--;
    while (i > 0) {
      dp.push_back(num[i - 1] - num[i]);
      i -= 2;
    }
    sum = 0;
    for (auto &x : dp) {
      sum += x;
      cas = max(sum, cas);
      if (sum < 0) sum = 0;
    }
    cout << ans + cas << "\n";
  }
}

### Prompt

Please formulate a Cpp solution 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>
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
using namespace std;
const long long maxn = 1e3 + 7;
const double eps = 1e-10;
const long long mod = 1e9 + 7;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> num(n);
    long long ans = 0;
    for (long long i = 0; i < (long long)(n); i++) {
      cin >> num[i];
      if ((i & 1) == 0) ans += num[i];
    }
    long long sum = 0;
    long long cas = 0;
    vector<long long> dp;
    long long i = 0;
    while (i + 1 < n) {
      dp.push_back(num[i + 1] - num[i]);
      i += 2;
    }
    for (auto &x : dp) {
      sum += x;
      cas = max(sum, cas);
      if (sum < 0) sum = 0;
    }
    dp.clear();
    i = n - 1;
    if (i & 1) i--;
    while (i > 0) {
      dp.push_back(num[i - 1] - num[i]);
      i -= 2;
    }
    sum = 0;
    for (auto &x : dp) {
      sum += x;
      cas = max(sum, cas);
      if (sum < 0) sum = 0;
    }
    cout << ans + cas << "\n";
  }
}

```