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;
const long long N = 100005;
const long long M = 1000000007;
void dfs(long long st, vector<vector<long long> > &v, vector<bool> &vis,
         vector<long long> &par) {
  if (vis[st]) return;
  vis[st] = true;
  for (long long i = 0; i < v[st].size(); i++) {
    if (vis[v[st][i]]) continue;
    par[v[st][i]] = st;
    dfs(v[st][i], v, vis, par);
  }
  return;
}
long long pow1(long long a, long long b, long long mod) {
  int res = 1;
  if (b < 0) b += mod - 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return res % mod;
}
void trav(vector<vector<long long> > &a,
          vector<pair<long long, long long> > &dp, long long cnt, long long x,
          long long y, long long n, long long m,
          map<pair<long long, long long>, bool> &ma) {
  if (x > (n - 1) || y > (m - 1)) return;
  if (ma[make_pair(x, y)]) return;
  ma[make_pair(x, y)] = true;
  if (a[x][y] == 0)
    dp[cnt].first++;
  else
    dp[cnt].second++;
  trav(a, dp, cnt + 1, x + 1, y, n, m, ma);
  trav(a, dp, cnt + 1, x, y + 1, n, m, ma);
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long no = 1;
  cin >> no;
  for (long long w = 0; w < no; w++) {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    vector<long long> od;
    vector<long long> ev;
    for (long long i = 1; i < n; i += 2) {
      od.push_back(a[i] - a[i - 1]);
    }
    for (long long i = 2; i < n; i += 2) {
      ev.push_back(a[i - 1] - a[i]);
    }
    long long sum = 0;
    for (long long i = 0; i < n; i += 2) {
      sum += a[i];
    }
    long long ma = 0;
    long long s2 = 0;
    for (long long i = 0; i < od.size(); i++) {
      s2 += od[i];
      if (s2 < 0) s2 = 0;
      ma = max(ma, s2);
    }
    s2 = 0;
    for (long long i = 0; i < ev.size(); i++) {
      s2 += ev[i];
      if (s2 < 0) s2 = 0;
      ma = max(ma, s2);
    }
    sum += ma;
    cout << sum << endl;
  }
  return 0;
}
