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 VI = vector <int>;
using VVI = vector <VI>;
inline int getBlockStart(int npos, int bakPos) {
    int nextBlock = (bakPos+1-2)/3 * 3 + 2;
    return npos < 0 ? nextBlock : max(nextBlock, (npos-2)/3*3 + 2);
}
inline int checkEqual(VI& A, int i, int j, int k){
    return A[i] == A[k] && A[i] == A[j];
}
inline void solve(VVI& dp, VI& A, int i, int j, int s) {
    if (s <= j && s+3 > j) s += 3;
    if (s <= j || j <= i) return ;
    int plus = s < A.size() ? checkEqual(A, i, j, s) : 0;
    dp[A.size()][A.size()] = max(dp[A.size()][A.size()], dp[i][j] + plus);
    if (s+2 < A.size()) {
        dp[i][s+0] = max(dp[i][s+0], dp[i][j] + checkEqual(A, j, s+1, s+2));
        dp[i][s+1] = max(dp[i][s+1], dp[i][j] + checkEqual(A, j, s+0, s+2));
        dp[i][s+2] = max(dp[i][s+2], dp[i][j] + checkEqual(A, j, s+0, s+1));
        dp[j][s+0] = max(dp[j][s+0], dp[i][j] + checkEqual(A, i, s+1, s+2));
        dp[j][s+1] = max(dp[j][s+1], dp[i][j] + checkEqual(A, i, s+0, s+2));
        dp[j][s+2] = max(dp[j][s+2], dp[i][j] + checkEqual(A, i, s+0, s+1));
        dp[s+0][s+1] = max(dp[s+0][s+1], dp[i][j] + checkEqual(A, i, j, s+2));
        dp[s+0][s+2] = max(dp[s+0][s+2], dp[i][j] + checkEqual(A, i, j, s+1));
        dp[s+1][s+2] = max(dp[s+1][s+2], dp[i][j] + checkEqual(A, i, j, s+0));
    }
}
int main() {
    int N;
    cin >> N;
    VI A;
    int ans = 0;
    for (int i = 0; i <= N; ++i) {
        int upper = 2 + (i > 0) - 2*(i == N);
        for (int j = 0; j < upper; ++j) {
            int v;
            cin >> v;
            A.push_back(v);
        }
        while (A.size() >= 3) {
            int aSize = A.size();
            if (A[aSize-1] == A[aSize-2] && A[aSize-1] == A[aSize-3]) {
                A.pop_back();
                A.pop_back();
                A.pop_back();
                ++ans;
            } else {
                break;
            }
        }
    }
    VVI dp(A.size()+1, VI(A.size()+1, -1));
    VI curPos(N+1, -1);
    VI nextPos(A.size() + 1);
    for (int p = A.size()-1; p >= 0; --p) {
        nextPos[p] = curPos[A[p]];
        curPos[A[p]] = p;
    }
    dp[0][1] = 0;
    for (int i= 0; i < A.size(); ++i) {
       for (int j = i+1; j < A.size(); ++j) {
            if (dp[i][j] == -1) {
                continue;
            }
            solve(dp, A, i, j, getBlockStart(j+1, j));
            solve(dp, A, i, j, getBlockStart(nextPos[i], j));
            solve(dp, A, i, j, getBlockStart(nextPos[j], j));
       }
    }
    if (!A.empty())
        cout << ans + dp[A.size()][A.size()] << endl;
	else
        cout << ans << endl;
    return 0;
}