Three Blocks Palindrome (hard version) CodeForces 1335E2

The only difference between easy and hard versions is constraints.

You are given a sequence aa consisting of nn positive integers.

Let's define a three blocks palindrome as the sequence, consisting of at most two distinct elements (let these elements are aa and bb, aa can be equal bb) and is as follows: [a,a,…,ax,b,b,…,by,a,a,…,ax][a,a,…,a⏟x,b,b,…,b⏟y,a,a,…,a⏟x]. There x,yx,y are integers greater than or equal to 00. For example, sequences [][], [2][2], [1,1][1,1], [1,2,1][1,2,1], [1,2,2,1][1,2,2,1] and [1,1,2,1,1][1,1,2,1,1] are three block palindromes but [1,2,3,2,1][1,2,3,2,1], [1,2,1,2,1][1,2,1,2,1] and [1,2][1,2] are not.

Your task is to choose the maximum by length subsequence of aa that is a three blocks palindrome.

You have to answer tt independent test cases.

Recall that the sequence tt is a a subsequence of the sequence ss if tt can be derived from ss by removing zero or more elements without changing the order of the remaining elements. For example, if s=[1,2,1,3,1,2,1]s=[1,2,1,3,1,2,1], then possible subsequences are: [1,1,1,1][1,1,1,1], [3][3] and [1,2,1,3,1,2,1][1,2,1,3,1,2,1], but not [3,2,3][3,2,3] and [1,1,1,1,2][1,1,1,1,2].

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases. Then tt test cases follow.

The first line of the test case contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2001≤ai≤200), where aiai is the ii-th element of aa. Note that the maximum value of aiai can be up to 200200.

It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105).

Output

For each test case, print the answer — the maximum possible length of some subsequence of aa that is a three blocks palindrome.

Example

Input

6
8
1 1 2 2 3 2 1 1
3
1 3 3
4
1 10 10 1
1
26
2
2 1
3
1 1 1

Output

7
2
4
1
1
3

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll inf = 1e18;
const int mod = 1000000007;
const int mx = 100; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
#define clr(a) memset(a, 0, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pair
void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
int  m, n,t,x,k,ans=0,sum=0;
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> t;
    while (t--) {
        cin >> n;
        vector<int>a(n);
        re(i, 0, n) { cin >> a[i]; --a[i]; }
        const int M = 200;
        vector<vector<int>>pref(M, vector<int>(n + 1));
        vector<vector<int>>at(M);
        re(i, 0, M) {
            re(j, 0, n) {
                pref[i][j + 1] = pref[i][j] + (a[j] == i);
                if (a[j] == i)at[i].push_back(j);
            }
        }
        ans = 0;
        re(x, 0, M) {
            int u = pref[x][n];
            ans = max(ans, u);
            for (int c = 1; 2 * c <= u; c++) {
                int from = at[x][c - 1];
                int to = at[x][u - c];
                for (int y = 0; y < M; y++) {
                    if (x != y) {
                        ans = max(ans, 2 * c + pref[y][to] - pref[y][from]);
                    }
                }
            }
        }
        cout << ans << endl;
    }
}