#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
// constexpr int MOD = 998244353;
constexpr int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
constexpr int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;

bool is_kadomatsu(ll a, ll b, ll c) {
  return a != b && b != c && c != a && !(a < b && b < c) && !(a > b && b > c);
}

void solve() {
  int n; cin >> n;
  vector<int> a(n); REP(i, n) cin >> a[i];
  int not_k = 0;
  FOR(i, 2, n) not_k += !is_kadomatsu(a[i - 2], a[i - 1], a[i]);
  FOR(s, 2, n) {
    if (!is_kadomatsu(a[s - 2], a[s - 1], a[s])) {
      for (int j = s - 2; j <= s; ++j) REP(i, n) {
        set<int> st;
        FOR(x, max(j - 2, 2), min(j + 3, n)) st.emplace(x);
        FOR(x, max(i - 2, 2), min(i + 3, n)) st.emplace(x);
        int tmp = not_k;
        for (int e : st) tmp -= !is_kadomatsu(a[e - 2], a[e - 1], a[e]);
        swap(a[j], a[i]);
        for (int e : st) tmp += !is_kadomatsu(a[e - 2], a[e - 1], a[e]);
        if (tmp == 0) {
          cout << "Yes\n";
          return;
        }
        swap(a[j], a[i]);
      }
      cout << "No\n";
      return;
    }
  }
  assert(false);
}

int main() {
  int t; cin >> t;
  while (t--) solve();
  return 0;
}