Round #338 (Div. 2)

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本文探讨了一个有趣的编程问题：如何通过按不同的按钮来点亮所有圣诞灯泡。问题涉及到布尔逻辑和状态变化，提供了一种判断是否能成功点亮所有灯泡的方法。

Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?

If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.

Input

The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.

Each of the next n lines contains x_i (0 ≤ x_i ≤ m) — the number of bulbs that are turned on by the i-th button, and then x_i numbers y_ij (1 ≤ y_ij ≤ m) — the numbers of these bulbs.

Output

If it's possible to turn on all m bulbs print "YES", otherwise print "NO".

Sample test(s)

Input

Output

YES

Input

Output

NO

Note

In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.

代码：

#include <stdio.h>
#include <string.h>
int main()
{
    int n,m,i,k,h;
    
    scanf("%d %d\n",&n,&m);
    int arr[m];
    memset(arr,0,sizeof(arr));
    while(n>0)
    {
        scanf("%d\n",&k);
        
        for(i=0;i<k;i++)
        {
            scanf("%d",&h);
            if(h!=0)
            arr[h-1]=1;
        }
        n--;
    }
    
    for(i=0;i<m;i++)
        if(arr[i]==0)
        {
            n = 200;
            printf("NO\n");
break;
        }
    if(n!=200)
        printf("YES\n");

    return 0;
}