Absent-minded Masha got set of *n* cubes for her birthday.

At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural *x* such she can make using her new cubes all integers from 1 to *x*.

To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.

The number can't contain leading zeros. It's not required to use all cubes to build a number.

Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
In first line integer *n* is given (1 ≤ *n* ≤ 3) — the number of cubes, Masha got for her birthday.

Each of next *n* lines contains 6 integers *a*<sub class="lower-index">*i*</sub><sub class="lower-index">*j*</sub> (0 ≤ *a*<sub class="lower-index">*i*</sub><sub class="lower-index">*j*</sub> ≤ 9) — number on *j*-th face of *i*-th cube.
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.
3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7
87
3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9
98
3
0 1 2 3 4 5
0 1 2 3 4 5
0 1 2 3 4 5
5
3
1 2 3 7 8 9
9 8 7 1 2 3
7 9 2 3 1 8
3
1
5 2 2 5 6 7
0
1
7 6 5 8 9 0
0
1
2 5 9 6 7 9
0
1
6 3 1 9 4 9
1
1
1 9 8 3 7 8
1
2
1 7 2 0 4 3
5 2 3 6 1 0
7
2
6 0 1 7 2 9
1 3 4 6 7 0
4
2
8 6 4 1 2 0
7 8 5 3 2 1
8
2
0 8 6 2 1 3
5 2 7 1 0 9
3
2
0 9 5 7 6 2
8 6 2 7 1 4
2
3
5 0 7 6 2 1
2 7 4 6 1 9
0 2 6 1 7 5
2
3
0 6 2 9 5 4
3 8 0 1 6 9
6 9 0 1 5 2
6
3
5 6 2 9 3 5
5 4 1 5 9 8
4 4 2 0 3 5
6
3
0 1 9 1 0 8
9 9 3 5 6 2
9 3 9 9 7 3
3
3
2 5 7 4 2 7
1 5 5 9 0 3
8 2 0 1 5 1
5
1
4 6 9 8 2 7
0
1
5 3 8 0 2 6
0
1
7 9 5 0 4 6
0
1
4 0 9 6 3 1
1
1
7 9 2 5 0 4
0
1
0 7 6 3 2 4
0
1
9 8 1 6 5 7
1
1
7 3 6 9 8 1
1
1
3 9 1 7 4 5
1
1
8 6 0 9 4 2
0
1
8 2 7 4 1 0
2
1
8 3 5 4 2 9
0
1
0 8 7 1 3 2
3
1
6 2 8 5 1 3
3
1
6 0 7 5 4 8
0
1
6 2 8 4 5 1
2
1
4 3 8 9 2 3
0
1
8 1 9 2 9 7
2
1
3 7 7 6 4 2
0
1
1 4 5 7 0 5
1
2
6 6 4 7 9 0
2 1 2 8 6 4
2
2
5 3 2 9 8 2
0 7 4 8 1 8
5
2
5 7 4 2 1 9
2 2 7 1 1 8
2
2
9 3 3 6 7 2
6 2 9 1 5 9
3
2
2 0 5 7 0 8
4 5 1 5 4 9
2
2
2 6 8 1 3 1
2 1 3 8 6 7
3
2
4 3 8 6 0 1
4 7 1 8 9 0
1
2
0 2 9 1 8 5
0 7 4 3 2 5
5
2
1 7 6 9 2 5
1 6 7 0 9 2
2
2
0 2 9 8 1 7
6 7 4 3 2 5
9
2
3 6 8 9 5 0
6 7 0 8 2 3
0
2
5 1 2 3 0 8
3 6 7 4 9 2
9
2
7 8 6 1 4 5
8 6 4 3 2 5
8
2
2 3 5 1 9 6
1 6 8 7 3 9
3
2
1 7 8 6 0 9
3 2 1 7 4 9
4
2
2 4 0 3 7 6
3 2 8 7 1 5
8
2
6 5 2 7 1 3
3 7 8 1 0 9
3
2
5 8 4 7 1 2
0 8 6 2 4 9
2
2
8 0 6 5 1 4
7 1 0 8 3 4
1
2
2 3 9 1 6 7
2 5 4 3 0 6
7
3
9 4 3 0 2 6
7 0 5 3 3 9
1 0 7 4 6 7
7
3
3 8 5 1 5 5
1 5 7 2 6 9
4 3 4 8 8 9
9
3
7 7 2 5 3 2
3 0 0 6 4 4
1 2 1 1 9 1
7
3
8 1 6 8 6 8
7 0 2 5 8 4
5 2 0 3 1 9
32
3
2 7 4 0 7 1
5 5 4 9 1 4
2 1 7 5 1 7
2
3
4 4 5 0 6 6
7 1 6 9 5 4
5 0 4 0 3 9
1
3
9 4 3 3 9 3
1 0 3 4 5 3
2 9 6 2 4 1
6
3
3 8 3 5 5 5
3 0 1 6 6 3
0 4 3 7 2 4
8
3
4 1 0 8 0 2
1 5 3 5 0 7
7 7 2 7 2 2
5
3
8 1 8 2 7 1
9 1 9 9 4 7
0 0 9 0 4 0
2
3
4 6 0 3 9 2
8 6 9 0 7 2
6 9 3 2 5 7
0
3
5 1 2 9 6 4
9 0 6 4 2 8
4 6 2 8 3 7
10
3
9 3 1 8 4 6
6 9 1 2 0 7
8 9 1 5 0 3
21
3
7 1 3 0 2 4
2 4 3 0 9 5
1 9 8 0 6 5
65
3
9 4 6 2 7 0
3 7 1 9 6 4
6 1 0 8 7 2
4
3
2 7 3 6 4 5
0 2 1 9 4 8
8 6 9 5 4 0
10
3
2 6 3 7 1 0
9 1 2 4 7 6
1 4 8 7 6 2
4
3
5 4 8 1 6 7
0 9 3 5 8 6
2 4 7 8 1 3
21
3
7 2 1 3 6 9
0 3 8 4 7 6
1 4 5 8 7 0
21
3
8 6 0 5 4 9
1 8 5 3 9 7
7 4 5 1 6 8
1
1
0 1 2 3 4 5
5
3
0 1 1 2 2 3
4 5 6 7 8 9
3 4 5 6 7 8
9
2
0 1 2 3 4 5
6 7 8 9 1 2
29
3
0 1 2 3 4 5
6 7 8 9 1 2
3 4 5 6 7 8
98
3
0 1 1 2 2 3
4 5 6 7 8 9
3 4 5 6 7 1
19
2
0 1 2 3 4 5
6 7 8 9 6 6
9
2
0 1 2 3 4 5
4 5 6 7 8 9
9
2
1 8 9 1 1 0
2 3 4 5 6 7
9
2
0 1 2 3 4 5
9 8 7 6 5 4
9
3
2 3 4 5 6 7
3 4 5 6 7 8
9 1 2 3 4 5
9
3
1 1 2 3 4 5
6 7 8 9 0 2
3 4 5 6 7 8
10
3
1 1 1 0 2 3
4 5 6 7 8 9
0 0 0 0 0 0
10
3
0 1 2 2 4 5
6 7 8 9 0 1
3 3 4 5 6 7
21
3
2 0 1 3 4 5
6 7 8 9 1 1
3 4 5 6 6 7
19
3
1 1 1 1 1 1
0 2 3 4 5 6
7 8 9 2 3 4
10
2
0 1 2 3 4 5
6 6 6 7 8 9
9
3
3 4 5 6 8 9
1 1 1 1 1 1
1 2 4 5 7 0
19
# -*- coding: utf-8 -*-
"""
Spyder Editor

num=list(map(int,input().split()))
[n,h]=list(map(int,input().split()))
a=int(input())

n,m =map(int,input().split())
c=[int(cost) for cost in input().split()] #fast enough speed

import math  # math.sqrt(x)->square root of x. cf min() is built in.
import queue # q=queue.Queue() it's better to define this outside of for loop
q.empty() it's important to put parenthesis

print (answer, " ", end='') this prints without changing the line

print(2,3) ->output: 2 3 (there's a space between 2 and 3)
print(2,end='')
print(3,end='') -->output: 23

outside of the block, variables are still stored

"""

import math
import queue

n=int(input())
a=[]

for i in range(0,n):
    a.append([int(c) for c in input().split()])
 
if n==3:
    for p in range(1,1001):
        exist=False
        for i in range(0,6):
            for j in range(0,6):
                for k in range(0,6):
                    if 100*a[0][i]+10*a[1][j]+a[2][k]==p:
                        exist=True
                    if 100*a[0][i]+10*a[2][k]+a[1][j]==p:
                        exist=True
                    if 100*a[1][j]+10*a[0][i]+a[2][k]==p:
                        exist=True
                    if 100*a[1][j]+10*a[2][k]+a[0][i]==p:
                        exist=True
                    if 100*a[2][k]+10*a[1][j]+a[0][i]==p:
                        exist=True
                    if 100*a[2][k]+10*a[0][i]+a[1][j]==p:
                        exist=True
                    
                    if 10*a[0][i]+a[1][j]==p:
                        exist=True
                    if 10*a[0][i]+a[2][k]==p:
                        exist=True   
                    if 10*a[1][j]+a[0][i]==p:
                        exist=True
                    if 10*a[1][j]+a[2][k]==p:
                        exist=True
                    if 10*a[2][k]+a[0][i]==p:
                        exist=True
                    if 10*a[2][k]+a[1][j]==p:
                        exist=True
                        
                    if a[0][i]==p:
                        exist=True
                    if a[1][j]==p:
                        exist=True
                    if a[2][k]==p:
                        exist=True
        if not exist:
            break
    
    
elif n==2:
    for p in range(1,101):
        exist=False
        for i in range(0,6):
            for j in range(0,6):
                if 10*a[0][i]+a[1][j]==p:
                    exist=True
                if 10*a[1][j]+a[0][i]==p:
                    exist=True
                if a[0][i]==p:
                    exist=True
                if a[1][j]==p:
                    exist=True
                    
        if not exist:
            break
        
    
elif n==1:
    for p in range(1,11):
        exist=False
        for i in range(0,6):
            if a[0][i]==p:
                exist=True
        if not exist:
             break

print(p-1)
    