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
#include <bits/stdc++.h>
using namespace std;
vector<int> adj[200000 + 1];
int cnt = 0;
vector<int> cand;
bool prime[400003];
void SieveOfEratosthenes(int n) {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
int prob = 0;
void dfs(vector<int> adj[], int v, bool visited[], vector<int>& span) {
  visited[v] = true;
  span.push_back(v);
  int i;
  for (i = 0; i < adj[v].size(); i++) {
    if (!visited[adj[v][i]]) {
      dfs(adj, adj[v][i], visited, span);
    }
  }
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 998244353;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % 998244353;
    y = y >> 1;
    x = (x * x) % 998244353;
  }
  return res;
}
long long add(long long x, long long y) {
  x += y;
  while (x >= 998244353) x -= 998244353;
  while (x < 0) x += 998244353;
  return x;
}
long long mul(long long x, long long y) { return (x * 1ll * y) % 998244353; }
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long inv(long long x) { return binpow(x, 998244353 - 2); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
long long fact[200009];
void precalc() {
  fact[0] = 1;
  for (long long i = 1; i < 200009; i++) fact[i] = mul(fact[i - 1], i);
}
long long C(long long n, long long k) {
  return divide(fact[n], mul(fact[k], fact[n - k]));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(25);
  cerr << fixed << setprecision(10);
  int t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long A[n];
    long long i;
    long long ans = 0;
    for (i = 0; i < n; i++) {
      cin >> A[i];
      if (i % 2 == 0) ans += A[i];
    }
    long long maxi = 0;
    long long sum = 0;
    for (i = 0; i + 1 < n; i += 2) {
      sum += A[i + 1] - A[i];
      if (sum < 0) sum = 0;
      if (sum > maxi) maxi = sum;
    }
    sum = 0;
    for (i = 1; i + 1 < n; i += 2) {
      sum += A[i] - A[i + 1];
      if (sum < 0) sum = 0;
      if (sum > maxi) maxi = sum;
    }
    cout << ans + maxi << "\n";
  }
  return 0;
}
