【Codeforces】-615A-Bulbs（思维，模型：桶）

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A. Bulbs

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 xi (0 ≤ xi ≤ m) —
the number of bulbs that are turned on by the i-th button, and then xi numbers yij(1 ≤ yij ≤ m) —
the numbers of these bulbs.

Output

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

Examples

input

3 4
2 1 4
3 1 3 1
1 2

output

YES

input

3 3
1 1
1 2
1 1

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.

把按钮可以按亮的灯标记为1，循环判断已知灯里有没有0的，有就不可以。

#include<cstdio>
int main()
{
int n,m,z=1;
int a[110]={0},b[110];
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&b[i]);
while(b[i]--)
{
int x;
scanf("%d",&x);
a[x]=1;<span style="white-space:pre"> </span>//亮灯标记为1
}
}
for(int i=1;i<=m;i++)
{
if(a[i]==0)<span style="white-space:pre"> </span>//有不为1的即不亮，退出
{
z=0;<span style="white-space:pre"> </span>
break;
}
}
if(z)
printf("YES\n");
else
printf("NO\n");
return 0;
}