#include <bits/stdc++.h>
using namespace std;
int n;
int a[10][10]; //a[i][1] ~ a[i][5]
int now[10]; //当前数的分解
int d[10], dCnt; //每一位需要拨动的范围, d 非 0 的个数
//检查 x 与 a[idx] 是否匹配
bool check(int x, int idx)
{
    now[1] = x;
    for(int i = 2; i <= 5; i++)
    {    
        now[i] = now[i - 1] / 10;
        now[i - 1] %= 10;
    }
    dCnt = 0;
    for(int i = 1; i <= 5; i++)
    {
        d[i] = now[i] - a[idx][i];
        if (d[i] < 0)
            d[i] += 10;
        if (d[i])
            dCnt ++;
    }
    if (dCnt == 0 || dCnt > 2)
        return false;
    if (dCnt == 1)
        return true;
    //两个不为 0 的数
    for (int i = 1; i <= 4; i++)
        if (d[i] && d[i] == d[i + 1])
            return true;
    return false;
}
int main()
{
    freopen("lock.in", "r", stdin);
    freopen("lock.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= 5; j++)
            cin >> a[i][j];
    int ans = 0;
    for (int i = 0; i <= 99999; i++)
    {
        bool flag = true;
        for (int j = 1; j <= n; j++)
        {
            if (!check(i,j))
            {
                flag = false;
                break;
            }
        }
        if (flag)
            ans++;
    }
    cout << ans << "\n";
    return 0;
}