codeforces-887B

B. Cubes for Masha

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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 aij (0 ≤ aij ≤ 9) — number on j-th face of i-th cube.

Output

Print single integer — maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can't make even 1.

Examples

input

3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7

output

87

input

3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9

output

98

Note

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.

方法很笨，而且十分麻烦，希望以后找到更好的方法。

#include<stdio.h>
void main()
{
    int a[3][6];
    int arr[250];
    int sum, n, i, j, k;
    scanf("%d", &n);
    for (i = 0;i < n;i++)
        for (j = 0;j < 6;j++)
            scanf("%d", &a[i][j]);
    switch (n)
    {
    case 1:
    {
        sum = 1;
        for (i = 0;i < 6;i++)
        {
            if (a[0][i] == sum)
            {
                sum++;
                i = 0;
            }
        }
        printf("%d
",sum-1);
        break;
    }
    case 2:
    {
        sum = 1;
        k = 0;
        for (i = 0;i < 6;i++)
            for (j = 0;j < 6;j++)
            {
                arr[k] = a[0][i] * 10 + a[1][j];
                arr[k + 1] = a[1][i] * 10 + a[0][j];
                k += 2;
            }
        for (i = 0;i < 6;i++)
        {
            arr[k] = a[0][i];
            arr[k + 1] = a[1][i];
            k += 2;
        }
        for (i = 0;i < k - 1;i++)
        {
            if (arr[i] == sum)
            {
                sum++;
                i = 0;
            }
        }
        printf("%d
", sum - 1);
        break;
    }
    case 3:
    {
        sum = 1;
        k = 0;
        for(i=0;i<6;i++)
            for (j = 0;j < 6;j++)
            {
                arr[k] = a[0][i] * 10 + a[1][j];
                arr[k + 1] = a[0][i] * 10 + a[2][j];
                arr[k + 2] = a[1][i] * 10 + a[0][j];
                arr[k + 3] = a[1][i] * 10 + a[2][j];
                arr[k + 4] = a[2][i] * 10 + a[1][j];
                arr[k + 5] = a[2][i] * 10 + a[0][j];
                k += 6;
            }
        for (i = 0;i < 6;i++)
        {
            arr[k] = a[0][i];
            arr[k + 1] = a[1][i];
            arr[k + 2] = a[2][i];
            k += 3;
        }
        for (i = 0;i < k - 1;i++)
        {
            if (arr[i] == sum)
            {
                sum++;
                i = 0;
            }
        }
        printf("%d
", sum - 1);
        break;
    }
    }    
}