題意:和E1相同，但是資料範圍擴大了，字元的數量為200，長度為2e5，如果按照E1的做法，時間複雜度為2e5*2e5*200，肯定T。

思路:把每一個字元出現的位置存進vector中，遍歷每一個字元，從兩邊取下標l，r。x已經確定，然後再遍歷200個字元從而確定y。

　　雖然是三個for，但是前兩個for實際上時間複雜度為n，所以總時間複雜度為2e5 * 200.

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include 
 
<map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define
 
 FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define
 
 repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
int sum[210][maxn];
int n;
void clearr()
{
    rep(i,1,n){
        rep(j,1,200){
            sum[j][i] = 0;
        }
    }
    //rep(i,1,n)
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int _;
    scanf("%d",&_);
    while(_--)
    {
        vector<int>pos[210];
        cin>>n;
        clearr();
        rep(i,1,n){
            rep(j,1,200){
                sum[j][i] = sum[j][i-1];
            }
            int x;
            cin>>x;
            sum[x][i]++;
            pos[x].pb(i);
        }
        int ans = 0;
        rep(i,1,200)
            ans = max(ans,sum[i][n]);

        rep(i,1,200){
            int l = 0,r = 0;
            int LEN = pos[i].size();
            if(LEN <= 1) continue; 
            rep(j,0,LEN/2-1){
                l = pos[i][j];
                r = pos[i][LEN-1-j];
                int lenmid = 0;
                rep(j,1,200){
                    lenmid = max(lenmid,sum[j][r-1]-sum[j][l]);
                }
                ans = max(ans,lenmid+(j+1)*2);                
            }
        cout<<ans<<endl;
    }
}