#include <bits/stdc++.h>

using namespace std;
#pragma GCC optimize("Ofast")

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef long long ll;
#define int ll
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define pb push_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define mispertion ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define F first
#define S second
#define getlast(s) (*s.rbegin())
#define debg cout << "OK\n"

const ld PI = 3.1415926535;
const int N = 200 + 10;
const int M = 1e5 + 10;
int mod = 998244353;
const int infi = INT_MAX;
const ll infl = 1e16;
const int P = 2;

int mult(int a, int b){
    return a * 1LL * b % mod;
}

int sum(int a, int b){
    if(a + b >= mod)
        return a + b - mod;
    if(a + b < 0)
        return a + b + mod;
    return a + b;
}

int binpow(int a, int n){
    if (n == 0)
        return 1;
    if (n % 2 == 1){
        return mult(binpow(a, n - 1), a);
    }
    else{
        auto b = binpow(a, n / 2);
        return mult(b, b);
    }
}

int n, a[N], dp[N][N][N];

void solve(){
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    dp[1][1][0] = 1;
    dp[1][0][1] = 1;
    int ans = 0;
    for(int i = 1; i < n; i++){
        for(int as = 0; as <= i; as++){
            for(int de = 0; de <= i; de++){
                if(as == de) continue;
                //i + 1 >> as
                int nxt = as, ad = 0;
                if(as == 0 || a[i + 1] > a[as])
                    nxt = i + 1, ad = 1;
                dp[i + 1][nxt][de] = max(dp[i + 1][nxt][de], dp[i][as][de] + ad);
                //i + 1 >> de
                nxt = de, ad = 0;
                if(de == 0 || a[i + 1] < a[de])
                    nxt = i + 1, ad = 1;
                dp[i + 1][as][nxt] = max(dp[i + 1][as][nxt], dp[i][as][de] + ad);
            }
        }
    }
    for(int i = 0; i <= n; i++){
        for(int j = 0; j <= n; j++){
            ans = max(ans, dp[n][i][j]);
        }
    }
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= n; j++)
            for(int k = 0; k <= n; k++)
                dp[i][j][k] = 0;
    cout << ans << '\n';
}

signed main(){
    mispertion;
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int t = 1;
    cin >> t;
    while (t--){
        solve();
    }
    return 0;
}