CodeForces 441A Valera and Antique Items-优快云博客

本文链接：https://blog.youkuaiyun.com/geekdoe/article/details/50464941

本文介绍了一个帮助Valera在有限资金下从多个古董商中挑选最便宜古董的算法。通过扫描每个商家提供的古董及其起拍价，算法找出Valera能够承担的古董，并返回这些商家的编号。

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Description

Valera is a collector. Once he wanted to expand his collection with exactly one antique item.

Valera knows n sellers of antiques, thei-th of them auctionedk_i items. Currently the auction price of thej-th object of thei-th seller iss_ij. Valera gets on well with each of then 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 then 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 k_i(1 ≤ k_i ≤ 50) the number of items of thei-th seller. Then go k_i space-separated integerss_i1, s_i2, ..., s_{ik_i}(10⁴ ≤ s_ij ≤ 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 integersq₁, q₂, ..., q_p(1 ≤ q_i ≤ n) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellersin the increasing order.

Sample Input

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

Hint

In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a40000 item from the first seller, a20000 item from the second seller, and a10000 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.

#include<iostream>
#include<algorithm>
using namespace std;

int main(){
    int n,v,p,i,price,num[60],flag,x;
    while(cin>>n>>v){
        x=flag=0;
        for(i=1;i<=n;i++){
            cin>>p;
            while(p--){
                cin>>price;
                if(v>price){
                    flag=1;
                }
            }
            if(flag){
                num[x]=i;
                x++;
                flag=0;
            }
        }
        cout<<x<<endl;
        for(i=0;i<x;i++){
            if(i==x-1){
                cout<<num[i]<<endl;
            }
            else{
                cout<<num[i]<<" ";
            }
        }
    }
    return 0;
}