#include <bits/stdc++.h>
#define int long long
#define inf 0x3f3f3f3f
#define ll long long
#define pii pair<int, int>
#define tii tuple<int, int, int>
#define db double
#define all(a) a.begin(), a.end()
using namespace std;
const int maxn = 2e5 + 10;
const int mod = 998244353;
int arr[maxn];
struct BIT {
    vector<pii> t;
    int len;
    BIT(int n) {len = n + 5;t.resize(len); fill(all(t), pii(0, 0));}
    int lowbit(int x) {return x & -x;}
    void update(int i, pii x) {
        i++;
        for (int pos = i; pos < len; pos += lowbit(pos)) {
            if (t[pos].first == x.first) t[pos].second += x.second;
            else if (t[pos].first < x.first)
                t[pos] = x;
        }
    }
    pii query(int i) {
        i++;
        pii res = {0, 0};
        for (int pos = i; pos; pos -= lowbit(pos))
            if (t[pos].first == res.first) res.second += t[pos].second;
            else if (t[pos].first > res.first) res = t[pos];
        return res;
    }
};
int f[maxn], g[maxn]; //以 i 结尾的LIS长度和方案数
int F[maxn], G[maxn]; //以 i 结尾的LDS长度和方案数
int N[maxn], M[maxn]; //以 i 开头的LIS的长度和方案数
int N2[maxn], M2[maxn]; // 以 i 开头的LDS的长度和方案数
int h[maxn]; // 必选 i 位置时的LIS方案数
int H[maxn]; // 必选 i 位置时的LDS方案数
void solve() {
    int n; cin >> n;
    for (int i = 0; i <= n; i++)
        h[i] = H[i] = 0;
    for (int i = 1; i <= n; i++) cin >> arr[i];
    int LIS = 0, LDS = 0;
    BIT bit1(n);
    for (int i = 1; i <= n; i++) {
        auto [len, cnt] = bit1.query(arr[i] - 1);
        if (++len == 1) cnt = 1;
        f[i] = len;
        g[i] = cnt;
        bit1.update(arr[i], {f[i], g[i]});
        LIS = max(LIS, f[i]);
    }
    BIT bit2(n);
    for (int i = 1; i <= n; i++) {
        auto [len, cnt] = bit2.query(n - arr[i]);
        if (++len == 1) cnt = 1;
        F[i] = len;
        G[i] = cnt;
        bit2.update(n + 1 - arr[i], {F[i], G[i]});
        LDS = max(LDS, F[i]);
    }
    BIT bit3(n);
    for (int i = n; i; i--) {
        auto [len, cnt] = bit3.query(n - arr[i]);
        if (++len == 1) cnt = 1;
        N[i] = len;
        M[i] = cnt;
        bit3.update(n + 1 - arr[i], {N[i], M[i]});
    }
    for (int i = 1; i <= n; i++) {
        if (f[i] + N[i] - 1 != LIS) continue;
        h[i] = g[i] * M[i];
    }
    BIT bit4(n);
    for (int i = n; i; i--) {
        auto [len, cnt] = bit4.query(arr[i] - 1);
        if (++len == 1) cnt = 1;
        N2[i] = len;
        M2[i] = cnt;
        bit4.update(arr[i], {N2[i], M2[i]});
    }
    for (int i = 1; i <= n; i++) {
        if (F[i] + N2[i] - 1 != LDS) continue;
        H[i] = G[i] * M2[i];
    }
    int c1 = 0, c2 = 0;
    for (int i = 1; i <= n; i++)  {
        if (f[i] == LIS) c1 += g[i];
        if (F[i] == LDS) c2 += G[i];
    }
    int sum = c1 * c2;
    for (int i = 1; i <= n; i++)
        sum -= h[i] * H[i];
    if (!sum)  cout << LIS + LDS - 1 << "\n";
    else cout << LIS + LDS << "\n";
}
signed main() {
    // freopen("1.in", "r", stdin);
    // freopen("1.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
}