/*** author: yuji9511 ***/
#include <bits/stdc++.h>
// #include <atcoder/all>
// using namespace atcoder;
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
using vll = vector<ll>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i=(m);i<(n);i++)
#define rrep(i,m,n) for(ll i=(m);i>=(n);i--)
ostream& operator<<(ostream& os, lpair& h){ os << "(" << h.first << ", " << h.second << ")"; return os;}
#define printa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
void print() {}
template <class H,class... T>
void print(H&& h, T&&... t){cout<<h<<" \n"[sizeof...(t)==0];print(forward<T>(t)...);}

void solve(){
    ll N;
    cin >> N;
    vll a(N);
    rep(i,0,N) cin >> a[i];
    ll ans = 0;
    rep(k,0,3){
        vll dp(N+1, 0);
        rep(i,k+3,k+N+1){
            if((a[(i-3)%N] - a[(i-2)%N]) * (a[(i-2)%N] - a[(i-1)%N]) < 0 && a[(i-3)%N] != a[(i-1)%N]){
                dp[i-k] = dp[i-k-3] + a[(i-3)%N];
            }
            dp[i-k] = max(dp[i-k], dp[i-k-1]);
        }
        ans = max(ans, dp[N]);
    }

    print(ans);

}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll t;
    cin >> t;
    while(t--) solve();
}