Problem:
Score : 400 points

Problem Statement

On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M.

For CODE FESTIVAL 20XX held on this planet, N participants gathered from all over the planet.

The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}.

Two participants A and B can communicate with each other if and only if one of the following conditions is satisfied:

There exists a language that both A and B can speak.

There exists a participant X that both A and B can communicate with.

Determine whether all N participants can communicate with all other participants.

Constraints

2≦N≦10^5

1≦M≦10^5

1≦K_i≦M

(The sum of all K_i)≦10^5

1≦L_{i,j}≦M

L_{i,1}, L_{i,2}, ..., L_{i,{}K_i} are pairwise distinct.

Partial Score

200 points will be awarded for passing the test set satisfying the following: N≦1000, M≦1000 and (The sum of all K_i)≦1000.

Additional 200 points will be awarded for passing the test set without additional constraints.

Input

The input is given from Standard Input in the following format:

N M
K_1 L_{1,1} L_{1,2} ... L_{1,{}K_1}
K_2 L_{2,1} L_{2,2} ... L_{2,{}K_2}
:
K_N L_{N,1} L_{N,2} ... L_{N,{}K_N}

Output

If all N participants can communicate with all other participants, print YES. Otherwise, print NO.

Sample Input 1

4 6
3 1 2 3
2 4 2
2 4 6
1 6

Sample Output 1

YES

Any two participants can communicate with each other, as follows:

Participants 1 and 2: both can speak language 2.

Participants 2 and 3: both can speak language 4.

Participants 1 and 3: both can communicate with participant 2.

Participants 3 and 4: both can speak language 6.

Participants 2 and 4: both can communicate with participant 3.

Participants 1 and 4: both can communicate with participant 2.

Note that there can be languages spoken by no participant.

Sample Input 2

4 4
2 1 2
2 1 2
1 3
2 4 3

Sample Output 2

NO

For example, participants 1 and 3 cannot communicate with each other.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

typedef struct {
	int N;
	int *u;
	int *u_rank;
}union_find;

union_find *make_union_find(int N){
	int i;
	union_find *uf = (union_find *)malloc(sizeof(union_find));
	uf->N = N;
	uf->u = (int *)malloc(sizeof(int) * N);
	uf->u_rank = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		uf->u[i] = i;
		uf->u_rank[i] = 1;
	}
	return uf;
}

int root_uf(int x, union_find *uf){
	int *u = uf->u;
	if(u[x] == x){
		return x;
	}
	else{
		u[x] = root_uf(u[x], uf);
		return u[x];
	}
}

void combine_uf(int x, int y, union_find *uf){
	int x_root = root_uf(x, uf);
	int y_root = root_uf(y, uf);
	int *u = uf->u;
	int *u_rank = uf->u_rank;
	if(x_root == y_root){
		return;
	}
	else if(u_rank[x_root] < u_rank[y_root]){
		u[x_root] = y_root;
		u_rank[y_root] += u_rank[x_root];
		u_rank[x_root] = 0;
	}
	else{
		u[y_root] = x_root;
		u_rank[x_root] += u_rank[y_root];
		u_rank[y_root] = 0;
	}
}

//xとyが同じ集合に属していれば1を,そうでなければ0を返す
int is_same_union_uf(int x, int y, union_find *uf){
	if(root_uf(x, uf) == root_uf(y, uf)){
		return 1;
	}
	else{
		return 0;
	}
}

//xが属する集合の要素数を返す
int rank_uf(int x, union_find *uf){
	return uf->u_rank[root_uf(x, uf)];
}

int main(){
	int N, M, K, L, r, i, j;
	scanf("%d%d", &N, &M);
	union_find *uf = make_union_find(N + M);
	for(i = 0; i < N; i++){
		scanf("%d", &K);
		for(j = 0; j < K; j++){
			scanf("%d", &L);
			L--;
			combine_uf(i, N + L, uf);
		}
	}
	r = root_uf(0, uf);
	for(i = 0; i < N; i++){
		if(root_uf(i, uf) != r){
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
}