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 PYTHON3):

Solution:
n = int(input())
A = list(map(int,input().split()))
dp = [[-float("INF")]*(n+1) for i in range(n+1)]
low = [-float("INF")]*(n+1)
m = 0
a,b = A[:2]
dp[a][b] = 0
dp[b][a] = 0
low[a] = 0
low[b] = 0
base = 0
for _ in range(1,n):
    a,b,c = A[3*_-1:3*_+2]
    q = []
    if a==b==c:
        base += 1
        continue
    m1 = m
    for j in range(3):
        q.append((a,b,max(m,dp[c][c]+1)))
        m1 = max(m1,dp[c][c]+1)
        for i in range(1,n+1):
            q.append((a,i,low[i]))
        if a==b:
            for i in range(1,n+1):
                q.append((c,i,dp[a][i]+1))
                m1 = max(m1,dp[a][i]+1)
        a,b,c = b,c,a
    for a,b,v in q:
        dp[a][b] = max(dp[a][b],v)
        dp[b][a] = max(dp[b][a],v)
        low[a] = max(low[a],v)  
        low[b] = max(low[b],v)      
    m = m1
print(max(m,dp[A[-1]][A[-1]]+1)+base)