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 Rust (1.15.1) (judge verdict: Accepted)

pub fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

pub fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

pub fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

struct UnionFind {
    parent: Option<usize>,
    size: u32,
}

impl Clone for UnionFind {
    fn clone(&self) -> Self {
        UnionFind {
            parent: self.parent.clone(),
            size: self.size,
        }
    }
}

fn root(a: usize, uf: &mut Vec<UnionFind>) -> usize {
    if let Some(v) = uf[a].parent {
        let tmp = root(v, uf);
        uf[a].parent = Some(tmp);
        tmp
    } else {
        a
    }
}

fn union(a: usize, b: usize, uf: &mut Vec<UnionFind>) {
    let ar = root(a, uf);
    let br = root(b, uf);
    if ar == br {
        return;
    }
    if uf[ar].size >= uf[br].size {
        uf[ar].size += uf[br].size;
        uf[br].parent = Some(ar);
    } else {
        uf[br].size += uf[ar].size;
        uf[ar].parent = Some(br);
    }
}

use std::collections::BTreeSet;

fn main() {
    let nm: Vec<usize> = read_vec();
    let n = nm[0];
    let m = nm[1];
    let mut uf = vec![
        UnionFind {
            parent: None,
            size: 1
        };
        m
    ];
    let mut s = BTreeSet::new();
    for _ in 0..n {
        let klv: Vec<usize> = read_vec();
        let k = klv[0];
        for i in 1..k {
            let l1 = klv[i] - 1;
            let l2 = klv[i + 1] - 1;
            union(l1, l2, &mut uf);
            s.insert(l1);
        }
        s.insert(klv[k] - 1);
    }
    let mut cnt = 0;
    for i in 0..m {
        if !s.contains(&i) {
            continue;
        }
        if uf[i].parent == None {
            cnt += 1;
        }
    }

    if cnt == 1 {
        println!("YES");
    } else {
        println!("NO");
    }
}