Title description
You may know that Euclid was a mathematician. Well, as it turns out, Morpheus knew it too. So when he wanted to play a mean trick on Euclid, he sent him an appropriate nightmare.
In his bad dream Euclid has a set S of n m-dimensional vectors over the Z2 field and can perform vector addition on them. In other words he has vectors with m coordinates, each one equal either 0 or 1. Vector addition is defined as follows: let u+v=w, then wi=(ui+vi)mod2.
Euclid can sum any subset of S and archive another m-dimensional vector over Z2. In particular, he can sum together an empty subset; in such a case, the resulting vector has all coordinates equal 0.
Let T be the set of all the vectors that can be written as a sum of some vectors from S. Now Euclid wonders the size of T and whether he can use only a subset S′ of S to obtain all the vectors from T. As it is usually the case in such scenarios, he will not wake up until he figures this out. So far, things are looking rather grim for the philosopher. But there is hope, as he noticed that all vectors in S have at most 2 coordinates equal 1.
Help Euclid and calculate |T|, the number of m-dimensional vectors over Z2 that can be written as a sum of some vectors from S. As it can be quite large, calculate it modulo 109+7. You should also find S′, the smallest such subset of S, that all vectors in T can be written as a sum of vectors from S′. In case there are multiple such sets with a minimal number of elements, output the lexicographically smallest one with respect to the order in which their elements are given in the input.
Consider sets A and B such that |A|=|B|. Let a1,a2,…a|A| and b1,b2,…b|B| be increasing arrays of indices elements of A and B correspondingly. A is lexicographically smaller than B iff there exists such i that aj=bj for all j<i and ai<bi.
Input
In the first line of input, there are two integers n, m (1≤n,m≤5⋅105) denoting the number of vectors in S and the number of dimensions.
Next n lines contain the description of the vectors in S. In each of them there is an integer k (1≤k≤2) and then follow k distinct integers x1,…xk (1≤xi≤m). This encodes an m-dimensional vector having 1s on coordinates x1,…xk and 0s on the rest of them.
Among the n vectors, no two are the same.
Output
In the first line, output two integers: remainder modulo 109+7 of |T| and |S′|. In the second line, output |S′| numbers, indices of the elements of S′ in ascending order. The elements of S are numbered from 1 in the order they are given in the input.
Examples
input
3 2
1 1
1 2
2 2 1
output
4 2
1 2
input
2 3
2 1 3
2 1 2
output
4 2
1 2
input
3 5
2 1 2
1 3
1 4
output
8 3
1 2 3
Note
In the first example we are given three vectors:
10
01
11
It turns out that we can represent all vectors from our 2-dimensional space using these vectors:
00 is a sum of the empty subset of above vectors;
01=11+10, is a sum of the first and third vector;
10=10, is just the first vector;
11=10+01, is a sum of the first and the second vector.
Hence, T={00,01,10,11}. We can choose any two of the three vectors from S and still be able to obtain all the vectors in T. In such a case, we choose the two vectors which appear first in the input. Since we cannot obtain all vectors in T using only a single vector from S, |S′|=2 and S′={10,01} (indices 1 and 2), as set {1,2} is lexicographically the smallest. We can represent all vectors from T, using only vectors from S′, as shown below:
00 is a sum of the empty subset;
01=01 is just the second vector;
10=10 is just the first vector;
11=10+01 is a sum of the first and the second vector.
General idea
Given n m-dimensional vectors, these vectors can be added modulo 2. Only two positions of these n vectors are 1 at most, and all other positions are 0.
How many kinds of vectors can be obtained by adding modulo 2 to these n vectors (this vector set is T), and which of the n vectors are required at least, the set T can also be obtained. And give the number of this group of vectors (if there are more than one, the solution with the smallest lexicographical order).
Topic analysis
Because each vector has at most two positions (assuming the numbers in these two positions are x and y respectively), we can connect an edge to x and y (because there is only one position with 1 in the vector , So we can set a virtual point and connect an edge between the position and the virtual point).
After this connection, a graph can be formed. If there are a edges in the graph that form a ring, then there are only a-1 valid edges (that is, a-1 valid vectors) in this ring .
Then the final number of effective vectors is: the number of rings in the m-graph.
Therefore, the number of vector types T that can be obtained is: 2 the number of valid vectors .
The problem is transformed to find how many rings there are in the graph. This problem can be solved by using union search.
code show as below
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <bitset>
#include <algorithm>
#define LL long long
#define PII pair<int,int>
#define x first
#define y second
using namespace std;
const int N=5e5+5,mod=1e9+7;
int p[N];
vector<int> ans;
int find(int x) //并查集模板
{
if(p[x]!=x) p[x]=find(p[x]);
return p[x];
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=m+1;i++) p[i]=i;
for(int i=1;i<=n;i++)
{
int k;
scanf("%d",&k);
int x,y=m+1; //虚拟点设为m+1
scanf("%d",&x);
if(k==2) scanf("%d",&y);
x=find(x),y=find(y); //找到x和y所在的集合
if(x!=y)
{
ans.push_back(i); //如果x和y不在同一集合，那么说明它为有效向量，用ans记录
p[x]=y; //并将其加入同一集合中去
}
}
int cnt=1;
for(int i=0;i<ans.size();i++) cnt=(cnt<<1)%mod; //T中的个数即为2^有效向量的数量
printf("%d %d\n",cnt,ans.size());
for(int i=0;i<ans.size();i++) //输出这些有效向量的编号
printf("%d ",ans[i]);
return 0;
}