#include <bits/stdc++.h>
using namespace std;
const long long inf = 2e18;
const long long mod = 1e9 + 7;
const long double pi = 3.141592653589793238462643383279502884;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void print(long long a[], long long n) {
  for (long long i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
long long power(long long x, long long y) {
  if (y <= 0) return 1;
  long long ans = 1;
  x %= mod;
  while (y) {
    if (y & 1) ans = (x * ans) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return ans;
}
long long modInverse(long long n) { return power(n, mod - 2); }
inline long long mul(long long a, long long b) { return (a * b) % mod; }
inline long long sub(long long a, long long b) {
  long long c = a - b;
  if (c < 0) c += mod;
  return c;
}
inline long long add(long long a, long long b) {
  long long c = a + b;
  if (c >= mod) c -= mod;
  return c;
}
inline long long divi(long long a, long long b) {
  return mul(a, modInverse(b));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(10);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    long long ans = 0, ans1 = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) ans += a[i];
    }
    long long e = 0, o = 0;
    long long em = INT_MAX, om = 0;
    long long ma = INT_MIN;
    for (int i = 0; i < n - 1; i += 2) {
      em = min(em, o - e);
      o += a[i + 1];
      e += a[i];
      ma = max(ma, o - e - em);
    }
    ans1 = max(ans, ans + ma);
    e = 0, o = 0;
    em = INT_MAX;
    for (int i = 1; i < n; i += 2) {
      em = min(em, o - e);
      e += a[i + 1];
      o += a[i];
      ma = max(ma, o - e - em);
    }
    cout << max(ans1, ans + ma) << "\n";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
const long long inf = 2e18;
const long long mod = 1e9 + 7;
const long double pi = 3.141592653589793238462643383279502884;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void print(long long a[], long long n) {
  for (long long i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
long long power(long long x, long long y) {
  if (y <= 0) return 1;
  long long ans = 1;
  x %= mod;
  while (y) {
    if (y & 1) ans = (x * ans) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return ans;
}
long long modInverse(long long n) { return power(n, mod - 2); }
inline long long mul(long long a, long long b) { return (a * b) % mod; }
inline long long sub(long long a, long long b) {
  long long c = a - b;
  if (c < 0) c += mod;
  return c;
}
inline long long add(long long a, long long b) {
  long long c = a + b;
  if (c >= mod) c -= mod;
  return c;
}
inline long long divi(long long a, long long b) {
  return mul(a, modInverse(b));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(10);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    long long ans = 0, ans1 = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) ans += a[i];
    }
    long long e = 0, o = 0;
    long long em = INT_MAX, om = 0;
    long long ma = INT_MIN;
    for (int i = 0; i < n - 1; i += 2) {
      em = min(em, o - e);
      o += a[i + 1];
      e += a[i];
      ma = max(ma, o - e - em);
    }
    ans1 = max(ans, ans + ma);
    e = 0, o = 0;
    em = INT_MAX;
    for (int i = 1; i < n; i += 2) {
      em = min(em, o - e);
      e += a[i + 1];
      o += a[i];
      ma = max(ma, o - e - em);
    }
    cout << max(ans1, ans + ma) << "\n";
  }
  return 0;
}

```