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;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first != b.first)
    return a.first < b.first;
  else
    return a.second > b.second;
}
long long ex(long long n, long long ans) {
  if (n == 1)
    return ans;
  else if (n % 2)
    return (ans % ((1000000007)) * ex(n / 2, (ans * ans) % (1000000007))) %
           (1000000007);
  else
    return ex(n / 2, (ans * ans) % (1000000007));
}
long long kmp(string x, long long m) {
  vector<long long> pi(m, 0);
  pi[0] = 0;
  long long j = 0;
  for (long long i = 1; i < m; i++) {
    j = pi[i - 1];
    while (j > 0 && x[i] != x[j]) j = pi[j - 1];
    if (x[j] == x[i]) j++;
    pi[i] = j;
  }
  return j;
}
string kmp1(string x, long long m) {
  vector<long long> pi(m, 0);
  pi[0] = 0;
  long long ma = 0;
  long long j = 0;
  long long pos = 0;
  for (long long i = 1; i < m; i++) {
    j = pi[i - 1];
    while (j > 0 && x[i] != x[j]) j = pi[j - 1];
    if (x[j] == x[i]) j++;
    pi[i] = j;
    if (x[i] == '#') pos = i;
  }
  long long p = pos;
  for (long long i = (pos); i < (m); ++i) {
    if (pi[i] > ma) {
      ma = pi[i];
      p = i;
    }
  }
  return x.substr(p - ma + 1, ma);
}
void dfs(vector<long long> adj[], long long fl[], long long i, long long& cn) {
  fl[i] = 1;
  cn++;
  for (long long j = (0); j < (adj[i].size()); ++j) {
    if (!fl[adj[i][j]]) {
      dfs(adj, fl, adj[i][j], cn);
    }
  }
}
int isp(long long x) {
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
struct Compare {
  bool operator()(long long const& p1, long long const& p2) {
    if (p1 < p2) {
      return true;
    }
    return false;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t1 = 1;
  cin >> t1;
  while (t1--) {
    long long n;
    cin >> n;
    long long a[n], s = 0;
    for (long long i = (0); i < (n); ++i) {
      cin >> a[i];
      if (i % 2 == 0) s += a[i];
    }
    long long p = s, ma = s;
    long long z = 0;
    for (long long i = (0); i < (n - 1); ++i) {
      p = p - a[i] + a[i + 1];
      z = z - a[i] + a[i + 1];
      if (z < 0) p = s, z = 0;
      ma = max(p, ma);
      i++;
    }
    p = s;
    z = 0;
    for (long long i = (1); i < (n - 1); ++i) {
      p = p + a[i] - a[i + 1];
      z = z + a[i] - a[i + 1];
      if (z < 0) p = s, z = 0;
      ma = max(p, ma);
      i++;
    }
    cout << (ma) << " ";
    cout << "\n";
    ;
  }
}
