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;
template <typename Arg1>
void prn(bool space, Arg1&& arg1) {
  cout << arg1;
  if (space)
    cout << " ";
  else
    cout << "\n";
}
template <typename Arg1, typename... Args>
void prn(bool space, Arg1&& arg1, Args&&... args) {
  prn(space, arg1);
  prn(space, args...);
}
const long long mod = 1000000007;
const long long MOD = 998244353;
const long long inf = 1e18;
const long long MAX = 2e5 + 1;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
vector<long long> fac(1e6);
long long pwr(long long x, long long n) {
  x = x % mod;
  if (!n) return 1;
  if (n & 1)
    return mul(x, pwr(mul(x, x), (n - 1) / 2));
  else
    return pwr(mul(x, x), n / 2);
}
long long modinv(long long n) { return pwr(n, mod - 2); }
long long inv(long long i) {
  if (i == 1) return 1;
  return (MOD - (MOD / i) * inv(MOD % i) % MOD) % MOD;
}
long long ncr(long long n, long long r) {
  long long ans = 1;
  if (r != 0 && r != n) {
    ans = fac[n] * modinv(fac[r]) % MOD * modinv(fac[n - r]) % MOD;
  }
  return ans;
}
void pre() {
  fac[0] = 1;
  for (int i = 1; i < (int)fac.size(); i++) {
    fac[i] = (fac[i - 1] * i) % MOD;
  }
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long dp[n][3];
  memset(dp, 0, sizeof(dp));
  dp[0][0] = a[0];
  dp[0][1] = a[0];
  dp[0][2] = a[0];
  dp[1][0] = a[0];
  dp[1][1] = max(a[1], a[0]);
  dp[1][2] = max(a[1], a[0]);
  for (int i = 2; i < n; i++) {
    dp[i][0] = max(dp[i][0], dp[i - 1][0] + (i % 2 == 0 ? a[i] : 0));
    dp[i][1] = max(dp[i][1], max(dp[i - 2][0], dp[i - 2][1]) +
                                 (i % 2 == 0 ? a[i - 1] : a[i]));
    dp[i][2] = max(dp[i][2], max({dp[i - 1][0], dp[i - 1][1], dp[i - 1][2]}) +
                                 (i % 2 == 0 ? a[i] : 0));
  }
  cout << max({dp[n - 1][0], dp[n - 1][1], dp[n - 1][2]}) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
