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:
/*input
2
1 2 1 2 2 1
*/

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;
void chmax(int& a, int b) { a = max(a, b);}

int dp[2005][2005];

int main() {
    int n;
    cin >> n;
    int n3 = n*3;
    vector<int> A(n3);
    rep(i, n3) cin >> A[i], --A[i];
    int m = n+1;
    rep(i, m) rep(j, m) dp[i][j] = -1e9;
    auto upd = [&](int i, int j, int x) {
        chmax(dp[i][j], x);
        chmax(dp[j][i], x);
        chmax(dp[i][n], x);
        chmax(dp[n][i], x);
        chmax(dp[n][j], x);
        chmax(dp[j][n], x);
        chmax(dp[n][n], x);
    };
    upd(A[0], A[1], 0);

    int base = 0;
    for (int ai = 2; ai < n3-1; ai += 3) {
        int x = A[ai], y = A[ai+1], z = A[ai+2];
        if (x == y && y == z) {
            ++base;
            continue;
        }
        vector<tuple<int, int, int>> q;
        rep(ri, 3) {
            {
                rep(a, m) {
                    int now = dp[a][n];
                    if (y == z) chmax(now, dp[a][y]+1);
                    q.emplace_back(a, x, now);
                }
            }
            {
                int now = max(dp[n][n], dp[z][z]+1);
                q.emplace_back(x, y, now);
            }
            swap(x, y); swap(y, z);
        }

        for (auto t: q) {
            int i, j, x;
            tie(i, j, x) = t;
            upd(i, j, x);
        }
    }

    int ans = dp[n][n];
    int l = A[n3-1];
    chmax(ans, dp[l][l]+1);
    ans += base;
    cout << ans << endl;
}