Problem:
Score : 600 points

Problem Statement

Given is a directed graph G with N vertices and M edges.

The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.

It is guaranteed that the graph contains no self-loops or multiple edges.

Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.

Here the null graph is not considered as a subgraph.

Notes

For a directed graph G = (V, E), we call a directed graph G' = (V', E') satisfying the following conditions an induced subgraph of G:

V' is a (non-empty) subset of V.

E' is the set of all the edges in E that have both endpoints in V'.

Constraints

1 \leq N \leq 1000

0 \leq M \leq 2000

1 \leq A_i,B_i \leq N

A_i \neq B_i

All pairs (A_i, B_i) are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
:
A_M B_M

Output

If there is no induced subgraph of G that satisfies the condition, print -1.
Otherwise, print an induced subgraph of G that satisfies the condition, in the following format:

K
v_1
v_2
:
v_K

This represents the induced subgraph of G with K vertices whose vertex set is \{v_1, v_2, \ldots, v_K\}. (The order of v_1, v_2, \ldots, v_K does not matter.)
If there are multiple subgraphs of G that satisfy the condition, printing any of them is accepted.

Sample Input 1

4 5
1 2
2 3
2 4
4 1
4 3

Sample Output 1

3
1
2
4

The induced subgraph of G whose vertex set is \{1, 2, 4\} has the edge set \{(1, 2), (2, 4), (4, 1)\}. The in-degree and out-degree of every vertex in this graph are both 1.

Sample Input 2

4 5
1 2
2 3
2 4
1 4
4 3

Sample Output 2

-1

There is no induced subgraph of G that satisfies the condition.

Sample Input 3

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2

Sample Output 3

4
2
3
4
5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
int n,m,tot,id,head[N],dfn[N],low[N],nex[N<<1],to[N<<1];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
int top,s[N];
bool vv[N],vis[N];
bool tarjan(int u)
{
    dfn[u]=low[u]=++id;
    s[++top]=u;vis[u]=true;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(vv[v]) continue;
        if(!dfn[v])
        {
            if(tarjan(v)) return true;
            low[u]=min(low[u],low[v]);
        }
        else if(vis[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])
    {
        if(s[top]!=u) return true;
        top--;vis[u]=false;
    }
    return false;
}
int main()
{
    scanf("%d%d",&n,&m);
    while(m--)
    {
        int u,v;scanf("%d%d",&u,&v);add(u,v);
    }
    bool ff=true;
    for(int i=1;i<=n;i++)
        if(!dfn[i])
    {
        if(tarjan(i)){ff=false;break;}
    }
    for(int i=1;i<=n;i++)
    {
        vv[i]=true;
        id=0;top=0;
        memset(dfn,0,sizeof(dfn));
        memset(vis,false,sizeof(vis));
        memset(low,0,sizeof(low));
        bool flag=false;
        for(int j=1;j<=n;j++)
            if(!dfn[j]&&!vv[j])
        {
            if(tarjan(j)) {flag=true;break;}
        }
        if(!flag) vv[i]=false;
    }
    int ans=0,p[N];
    for(int i=1;i<=n;i++)
        if(!vv[i]) p[++ans]=i;
    if(ff) printf("-1\n");
    else
    {
        printf("%d\n",ans);
        for(int i=1;i<=ans;i++)
            printf("%d\n",p[i]);
    }
}