Codeforces Round #252(Div. 2) 441A. Valera and Antique Items 水题

A. Valera and Antique Items

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows n sellers
of antiques, the i-th of them auctioned ki items.
Currently the auction price of the j-th object of the i-th
seller issij.
Valera gets on well with each of the n sellers. He is perfectly sure that if he
outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only v units
of money. Help him to determine which of the n sellers he can make a deal with.

Input
The first line contains two space-separated integers n, v (1 ≤ n ≤ 50; 104 ≤ v ≤ 106) —
the number of sellers and the units of money the Valera has.
Then n lines
follow. The i-th line first contains integer ki (1 ≤ ki ≤ 50) the
number of items of the i-th seller. Then go ki space-separated
integers si1, si2, ..., siki (104 ≤ sij ≤ 106) —
the current prices of the items of the i-th seller.

Output
In the first line, print integer p —
the number of sellers with who Valera can make a deal.
In the second line print p space-separated
integers q1, q2, ..., qp (1 ≤ qi ≤ n) —
the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.

Examples

input

3 50000
1 40000
2 20000 60000
3 10000 70000 190000

output

3
1 2 3

input

3 50000
1 50000
3 100000 120000 110000
3 120000 110000 120000

output

0

Note
In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item
from the first seller, a20000 item from the second seller, and a 10000 item
from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him.
题意：RT
题解：
扫一遍，判断就是了。

/****************
*PID:441a div2
*Auth:Jonariguez
*****************
*/
#define lson k*2,l,m
#define rson k*2+1,m+1,r
#define rep(i,s,e) for(i=(s);i<=(e);i++)
#define For(j,s,e) For(j=(s);j<(e);j++)
#define sc(x) scanf("%d",&x)
#define In(x) scanf("%I64d",&x)
#define pf(x) printf("%d",x)
#define pfn(x) printf("%d\n",(x))
#define Pf(x) printf("%I64d",(x))
#define Pfn(x) printf("%I64d\n",(x))
#define Pc printf(" ")
#define PY puts("YES")
#define PN puts("NO")
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef int Ll;
Ll quick_pow(Ll a,Ll b,Ll MOD){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}

const int maxn=100000+10;
vector<int> ans;

int main()
{
int i,j,n,m;
while(scanf("%d%d",&n,&m)!=EOF){
int res=0;
ans.clear();
for(i=1;i<=n;i++){
int k,x;
scanf("%d%d",&k,&x);
int y;
for(j=1;j<k;j++){
sc(y);x=min(y,x);
}
if(m>x)
ans.push_back(i);
}
printf("%d\n",ans.size());
for(i=0;i<ans.size();i++)
printf("%d ",ans[i]);
puts("");
}
return 0;
}