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>
#pragma GCC optimize("O3")
#pragma GCC target("avx")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long INF = 1e18;
const long long NEG_INF = -1 * (1e18);
using namespace std;
void usaco(string prob) {
  freopen((prob + ".in").c_str(), "r", stdin);
  freopen((prob + ".out").c_str(), "w", stdout);
}
bool prime(int n) {
  if (n < 2) return false;
  for (int x = 2; x * x <= n; x++) {
    if (n % x == 0) return false;
  }
  return true;
}
vector<int> getFactors(int n) {
  vector<int> f;
  for (int x = 2; x * x <= n; x++) {
    while (n % x == 0) {
      f.push_back(x);
      n /= x;
    }
  }
  if (n > 1) f.push_back(n);
  return f;
}
void solve() {
  long long n, ans = 0, del1 = 0, del2 = 0, sum = 0, csum = 0, msum = INF;
  cin >> n;
  vector<long long> a(n);
  vector<long long> v1, v2;
  for (long long i = 0; i < (long long)(n); ++i) {
    cin >> a[i];
    if (i % 2 == 0) {
      sum += a[i];
    }
  }
  if (n == 1) {
    cout << a[0] << "\n";
    return;
  }
  for (long long i = 1; i < n; i += 2) {
    v1.push_back(a[i] - a[i - 1]);
  }
  for (long long i = 2; i < n; i += 2) {
    v2.push_back(a[i - 1] - a[i]);
  }
  vector<long long> tmp(n, 0);
  if (!v1.empty()) {
    tmp[0] = del1 = v1[0];
    for (long long i = 1; i <= (long long)(v1.size() - 1); ++i) {
      tmp[i] = max(v1[i], v1[i] + tmp[i - 1]);
      del1 = max(del1, tmp[i]);
    }
  }
  fill((tmp).begin(), (tmp).end(), 0);
  csum = 0, msum = INF;
  if (!v2.empty()) {
    tmp[0] = del2 = v2[0];
    for (long long i = 1; i <= (long long)(v2.size() - 1); ++i) {
      tmp[i] = max(v2[i], v2[i] + tmp[i - 1]);
      del2 = max(del2, tmp[i]);
    }
  }
  ans = sum + max({0LL, del1, del2});
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long TC = 1;
  cin >> TC;
  for (long long tt = 1; tt <= (long long)(TC); ++tt) {
    solve();
  }
  return 0;
}
