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>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1e3+3;
int n, m, scc;
int w[maxn][maxn];
int dfn[maxn], low[maxn], vis[maxn], Index;
int belong[maxn], cnt[maxn], outdeg[maxn];
int s[maxn], top;
vector<int> g[maxn], ss[maxn];

void read()
{
    cin >> n >> m;
    int u, v;
    for(int i = 0; i < m; ++i)
    {
        cin >> u >> v;
        g[u].push_back(v);
        w[u][v] = 1;
    }
}

void tarjan(int u)
{
    dfn[u] = low[u] = ++Index;
    s[top++] = u;
    for(int i = 0; i < g[u].size(); ++i)
    {
        int v = g[u][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(!vis[v])
            low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u])
    {
        ++scc;
        while(1)
        {
            int x = s[--top];
            vis[x] = 1;
            belong[x] = scc;
            ss[scc].push_back(x);
            ++cnt[scc];
            if(x == u)
                break;
        }
    }
}

int viss[maxn], p[maxn], pc = 0;
bool flag = 0;
void dfs(int t, int u)
{
    if(flag) return;
    p[pc++] = u;
    viss[u] = 1;
    for(int i = pc-1; i >= 0; --i)
    {
        int v = p[i];
        if(w[u][v] && viss[v])
        {
            int ans = 0, j;
            for(j = pc-1; j >= 0; --j)
                if(p[j] == v)
                {
                    ans = pc-j;
                    break;
                }
            cout << ans << endl;
            for(int k = j; k < pc; ++k)
                cout << p[k] << endl;
            flag = 1;
            return;
        }
    }
    for(int i = 0; i < g[u].size(); ++i)
    {
        int v = g[u][i];
        if(belong[v] != t)
            continue;
        dfs(t, v);
        if(flag)
            return;
    }
}

void solve()
{
    for(int i = 1; i <= n; ++i)
        if(!dfn[i])
            tarjan(i);
    
    int minscc = 0, mincnt = INF;
    for(int i = 1; i <= scc; ++i)
        if(cnt[i] > 1 && cnt[i] < mincnt)
        {
            minscc = i;
            mincnt = cnt[i];
        }
    if(!minscc)
    {
        cout << -1 << endl;
        return;
    }

    for(int i = 0; i < cnt[minscc]; ++i)
        if(!flag)
        {
            memset(vis, 0, sizeof(vis));
            memset(p, 0, sizeof(p));
            pc = 0;
            dfs(minscc, ss[minscc][i]);
        }
    if(!flag)
        cout << -1 << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    read();
    solve();
    return 0;
}