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;
typedef long long ll;
const int N = 3e5 + 10;

int n, m;
struct Edge {
    int to, next;
}e[N];
int cnt, head[N];
void addedge(int u, int v) {
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt++;
}
int sz, dfn[N], low[N], vis[N];
int scc;
vector<int> ans;
stack<int> sta;
void tarjan(int u) {
    dfn[u] = low[u] = ++sz;
    vis[u] = 1;
    sta.push(u);
    for(int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to;
        if(!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if(vis[v] && low[u] > dfn[v]) {
            low[u] = dfn[v];
        }
    }
    if(low[u] == dfn[u] && ans.empty()) {
        scc++;
        while(1) {
            int id = sta.top(); sta.pop();
            vis[id] = 0;
            if(low[id] == low[u]) {
                ans.push_back(id);
            }
            if(id == u) break;
        }
        if(ans.size() == 1) ans.clear();
    }
}
int main() {
    cin >> n >> m;
    memset(vis, 0, sizeof vis);
    while(!sta.empty()) sta.pop();
    scc = cnt = sz = 0;
    memset(head, -1, sizeof head);
    memset(dfn, 0, sizeof dfn);
    for(int i = 0; i < m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        addedge(u, v);
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i]) tarjan(i);
    }
    if(ans.empty()) cout << -1 << '\n';
    else {
        cout << ans.size() << '\n';
        for(int x : ans) {
            printf("%d\n", x);
        }
    }
    return 0;
}