Problem: We have 3N cards arranged in a row from left to right, where each card has an integer between 1 and N (inclusive) written on it. The integer written on the i-th card from the left is A_i.

You will do the following operation N-1 times:

* Rearrange the five leftmost cards in any order you like, then remove the three leftmost cards. If the integers written on those three cards are all equal, you gain 1 point.



After these N-1 operations, if the integers written on the remaining three cards are all equal, you will gain 1 additional point.

Find the maximum number of points you can gain.

Constraints

* 1 \leq N \leq 2000
* 1 \leq A_i \leq N

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_{3N}


Output

Print the maximum number of points you can gain.

Examples

Input

2
1 2 1 2 2 1


Output

2


Input

3
1 1 2 2 3 3 3 2 1


Output

1


Input

3
1 1 2 2 2 3 3 3 1


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define ALL(obj) (obj).begin(),(obj).end()
template <class T> void corner(bool flg, T hoge) {if (flg) {cout << hoge << endl; exit(0);}}
template <class T> inline void chmax(T& a, const T b){a=max(a,b);}

array<int,6000> A;
array<int,2000> B;
array<int,2000*2000> dp;

int main() {
    cin.tie(0);ios::sync_with_stdio(false);
    constexpr int inf = 12345678;
    int N; cin >> N;
    for(int i = 0; i < 3*N; ++i) cin >> A[i],A[i]--;
    corner(N==1,A[0]==A[1]&&A[1]==A[2]);

    for(int i=0;i<N*N;++i) dp[i]=-inf;
    for(int i=0;i<N;++i) B[i]=-inf;
    int ans=0,maxi=0;
    array<int,5> v;
    for(int i = 0; i < 5; ++i) v[i]=A[i];
    sort(ALL(v));
    do{
        int u = (v[2]==v[3]&&v[3]==v[4]);
        chmax(dp[v[0]*N+v[1]],u);
        chmax(maxi,u);
        chmax(B[v[0]],u);
        chmax(B[v[1]],u);
    } while(next_permutation(ALL(v)));

    for(int i=5; i+2 < 3*N; i+=3) {
        int& a=A[i];
        int& b=A[i+1];
        int& c=A[i+2];
        if(a==b&&b==c) {
            ans++;
            continue;
        }
        stack<pair<int,int>> vp;
        if(a==b) {
            for(int j=0;j<N;++j) {
                vp.emplace(c*N+j,max(dp[a*N+j]+1,B[j]));
            }
        }
        else {
            for(int j=0;j<N;++j) {
                vp.emplace(c*N+j,B[j]);
            }
        }
        if(b==c) {
            for(int j=0;j<N;++j) {
                vp.emplace(a*N+j,max(dp[b*N+j]+1,B[j]));
            }
        }
        else{
            for(int j=0;j<N;++j) {
                vp.emplace(a*N+j,B[j]);
            }
        }
        if(c==a) {
            for(int j=0;j<N;++j) {
                vp.emplace(b*N+j,max(dp[c*N+j]+1,B[j]));
            }
        }
        else{
            for(int j=0;j<N;++j) {
                vp.emplace(b*N+j,B[j]);
            }
        }
        {
            vp.emplace(b*N+c,max(dp[a*N+a]+1,maxi));
        }
        {
            vp.emplace(c*N+a,max(dp[b*N+b]+1,maxi));
        }
        {
            vp.emplace(a*N+b,max(dp[c*N+c]+1,maxi));
        }
        while(vp.size()) {
            auto p = vp.top(); vp.pop();
            int s=p.first/N,t=p.first%N,u=p.second;
            chmax(dp[s*N+t],u);
            chmax(dp[t*N+s],u);
            chmax(maxi,u);
            chmax(B[s],u);
            chmax(B[t],u);
        }
    }
    int& l=A[3*N-1];
    chmax(maxi,dp[l*N+l]+1);
    ans += maxi;
    cout << ans << endl;
    return 0;
}