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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author silviase
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CInterpretation solver = new CInterpretation();
        solver.solve(1, in, out);
        out.close();
    }

    static class CInterpretation {
        public void solve(int testNumber, Scanner in, PrintWriter out) {

            int n = in.nextInt();
            int m = in.nextInt();
            UnionFindTree uf = new UnionFindTree(n);
            ArrayList<Integer>[] lang = new ArrayList[m];

            for (int i = 0; i < m; i++) {
                lang[i] = new ArrayList<>();
            }

            // しゃべれるものを追加
            for (int i = 0; i < n; i++) {
                int k = in.nextInt();
                for (int j = 0; j < k; j++) {
                    lang[in.nextInt() - 1].add(i);
                }
            }

            // 各言語にunion
            for (int i = 0; i < m; i++) {
                for (int j = 1; j < lang[i].size(); j++) {
                    uf.union(lang[i].get(0), lang[i].get(j));
                }
            }


            // 判定
            for (int i = 1; i < n; i++) {
                if (!uf.same(0, i)) {
                    out.println("NO");
                    return;
                }
            }
            out.println("YES");

        }

    }

    static class UnionFindTree {
        private int[] parent;
        private int[] height;
        private int[] size;

        public UnionFindTree(int size) {
            this.parent = new int[size];
            this.height = new int[size];
            this.size = new int[size];

            for (int i = 0; i < size; i++) {
                makeSet(i);
            }
        }

        private void makeSet(int i) {
            parent[i] = i;
            height[i] = 0;
            size[i] = 1;
        }

        public int find(int x) {
            if (x != parent[x]) {
                parent[x] = find(parent[x]);
            }
            return parent[x];
        }

        public void union(int x, int y) {

            int rootX = find(x);
            int rootY = find(y);

            if (rootX == rootY) { //既に結合済み
                return;
            }

            if (height[rootX] > height[rootY]) {
                parent[rootY] = rootX;
                size[rootX] += size[rootY];

            } else if (height[rootX] < height[rootY]) {
                parent[rootX] = rootY;
                size[rootY] += size[rootX];

            } else {
                parent[rootY] = rootX;
                size[rootX] += size[rootY];
                height[rootX]++;
            }
        }

        public boolean same(int x, int y) {
            return find(x) == find(y);
        }

    }
}