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: Wrong Answer)


#include<bits/stdc++.h>
using namespace std;
/**
********************************************************************
******************* Author:Bisnu sarkar ****************************
********************************************************************
**/
#define ull            unsigned long long
#define ll             long long
#define pii            pair<int,int>
#define sit            set<int> :: iterator
#define vrit           vector<int> :: reverse iterator
#define ff             first
#define ss             second
#define endl           '\n';

int set_1(int n,int pos){return n = (n | (1<<pos));}
int reset_0(int n,int pos){return n= n & ~(1<<pos);}
bool check_bit(int n,int pos){return n = n & (1<<pos);}

int dx[8]= {1,0,-1,0,-1,-1,1,1};
int dy[8]= {0,1,0,-1,-1,1,-1,1};
const int N = (int) 2e3 + 5;
const int M = (int) 1e9 + 7;
const double pi=2* acos(0.0);
const double eps=1e-9;


vector<int> color;
vector<int> parent;
int cycle_start, cycle_end;

int n;
std::vector<int> adj[N];

bool dfs(int v) {
    color[v] = 1;
    for (int i=0;i<adj[v].size();++i) {
        int u=adj[v][i];
        if (color[u] == 0) {
            parent[u] = v;
            if (dfs(u))
                return true;
        } else if (color[u] == 1) {
            cycle_end = v;
            cycle_start = u;
            return true;
        }
    }
    color[v] = 2;
    return false;
}

void find_cycle() {
    color.assign(n, 0);
    parent.assign(n, -1);
    cycle_start = -1;

    for (int v = 0; v < n; v++) {
        if (color[v] == 0 && dfs(v))
            break;
    }

    if (cycle_start == -1) {
        cout << "-1" << endl;
    } else {
        vector<int> cycle;
        cycle.push_back(cycle_start);
        for (int v = cycle_end; v != cycle_start; v = parent[v])
            cycle.push_back(v);
       cycle.push_back(cycle_start);
        reverse(cycle.begin(), cycle.end());

        printf("%d\n", (int)cycle.size()-1);
       for(int i=0;i<(int)cycle.size()-1;++i){
        printf("%d\n", cycle[i]);
       }
       
    }
}

int main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL); 

    int t=1,te=0;
    // cin>>t;
    while(t--){
        int m;
        cin>>n>>m;

        for(int i=1;i<=m;++i){
            int u,v;
            cin>>u>>v;
            adj[u].push_back(v);
        }

      

        find_cycle();


        //printf("Case %d: ",++te);
    }
    return 0;
}