【codeforces 757C】Felicity is Coming!

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

It’s that time of the year, Felicity is around the corner and you can see people celebrating all around the Himalayan region. The Himalayan region has n gyms. The i-th gym has gi Pokemon in it. There are m distinct Pokemon types in the Himalayan region numbered from 1 to m. There is a special evolution camp set up in the fest which claims to evolve any Pokemon. The type of a Pokemon could change after evolving, subject to the constraint that if two Pokemon have the same type before evolving, they will have the same type after evolving. Also, if two Pokemon have different types before evolving, they will have different types after evolving. It is also possible that a Pokemon has the same type before and after evolving.

Formally, an evolution plan is a permutation f of {1, 2, …, m}, such that f(x) = y means that a Pokemon of type x evolves into a Pokemon of type y.

The gym leaders are intrigued by the special evolution camp and all of them plan to evolve their Pokemons. The protocol of the mountain states that in each gym, for every type of Pokemon, the number of Pokemon of that type before evolving any Pokemon should be equal the number of Pokemon of that type after evolving all the Pokemons according to the evolution plan. They now want to find out how many distinct evolution plans exist which satisfy the protocol.

Two evolution plans f1 and f2 are distinct, if they have at least one Pokemon type evolving into a different Pokemon type in the two plans, i. e. there exists an i such that f1(i) ≠ f2(i).

Your task is to find how many distinct evolution plans are possible such that if all Pokemon in all the gyms are evolved, the number of Pokemon of each type in each of the gyms remains the same. As the answer can be large, output it modulo 109 + 7.

Input

The first line contains two integers n and m (1 ≤ n ≤ 105, 1 ≤ m ≤ 106) — the number of gyms and the number of Pokemon types.

The next n lines contain the description of Pokemons in the gyms. The i-th of these lines begins with the integer gi (1 ≤ gi ≤ 105) — the number of Pokemon in the i-th gym. After that gi integers follow, denoting types of the Pokemons in the i-th gym. Each of these integers is between 1 and m.

The total number of Pokemons (the sum of all gi) does not exceed 5·105.

Output

Output the number of valid evolution plans modulo 109 + 7.

Examples

input

2 3

2 1 2

2 2 3

output

1

input

1 3

3 1 2 3

output

6

input

2 4

2 1 2

3 2 3 4

output

2

input

2 2

3 2 2 1

2 1 2

output

1

input

3 7

2 1 2

2 3 4

3 5 6 7

output

24

Note

In the first case, the only possible evolution plan is:

In the second case, any permutation of (1,  2,  3) is valid.

In the third case, there are two possible plans:

In the fourth case, the only possible evolution plan is:

【题目链接】:http://codeforces.com/contest/757/problem/C

【题解】



记录最后每种动物在哪些gym里面出现过;

如果2动物在3号gym里面出现两次,在4号gym里面出现3次;

定义

    vector <int> a[MAXM];

则a[2]={3,3,4,4,4};

如果3号动物和2号之间能互相进化

则必有a[2]=a[3]

即2号动物在哪个gym里面出现过,3号动物也要在那个gym里面出现;

且要求出现的次数相同;(所以a[x]里面可能有重复的数字表示它在这个

gym里面出现多次);

(如果符合这样的要求,则2和3能够转化,且也这有这样才能满足转化过后2和3在各个gym里面出现的次数都不变);

然后用一个sort把a[1..m]都排序一下;

a里面的内容也要排序;

这样相同的a就会在靠在一起;

那些相同的a能够互相转化;

每个相同的a构成的若干个集合设为Si

则每个集合的方案数为(Si)!

答案就是

∏((Si)!)

预处理出1..MAXM的阶乘;

那些没有出现的动物;显然可以任意转化;

所以也是N！



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 110;
const int MAXM = 1e6+100;
const LL MOD = 1e9+7;
int n,m;
vector <int> a[MAXM];
LL jc[MAXM];
int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    jc[0] = 1;
    for (LL i = 1;i <= 1000000;i++)
        jc[i] = (jc[i-1]*i)%MOD;
    rei(n);rei(m);
    rep1(i,1,n)
    {
        int num;
        rei(num);
        rep1(j,1,num)
            {
                int x;
                rei(x);
                a[x].pb(i);
            }
    }
    sort(a+1,a+1+m);
    LL ans = 1;
    rep1(i,1,m)
    {
        int j = i + 1;
        while (j<=m && a[j]==a[i]) j++;
        int len = j-i;
        ans = (ans * jc[len])%MOD;
        i = j-1;
    }
    cout << ans << endl;
    return 0;
}